wrong_submission_id
stringlengths 10
10
| problem_id
stringlengths 6
6
| user_id
stringlengths 10
10
| time_limit
float64 1k
8k
| memory_limit
float64 131k
1.05M
| wrong_status
stringclasses 2
values | wrong_cpu_time
float64 10
40k
| wrong_memory
float64 2.94k
3.37M
| wrong_code_size
int64 1
15.5k
| problem_description
stringlengths 1
4.75k
| wrong_code
stringlengths 1
6.92k
| acc_submission_id
stringlengths 10
10
| acc_status
stringclasses 1
value | acc_cpu_time
float64 10
27.8k
| acc_memory
float64 2.94k
960k
| acc_code_size
int64 19
14.9k
| acc_code
stringlengths 19
14.9k
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
s505636196
|
p02406
|
u801346721
| 1,000 | 131,072 |
Time Limit Exceeded
| 40,000 | 7,540 | 143 |
In programming languages like C/C++, a goto statement provides an unconditional jump from the "goto" to a labeled statement. For example, a statement "goto CHECK_NUM;" is executed, control of the program jumps to CHECK_NUM. Using these constructs, you can implement, for example, loops. Note that use of goto statement is highly discouraged, because it is difficult to trace the control flow of a program which includes goto. Write a program which does precisely the same thing as the following program (this example is wrtten in C++). Let's try to write the program without goto statements. void call(int n){ int i = 1; CHECK_NUM: int x = i; if ( x % 3 == 0 ){ cout << " " << i; goto END_CHECK_NUM; } INCLUDE3: if ( x % 10 == 3 ){ cout << " " << i; goto END_CHECK_NUM; } x /= 10; if ( x ) goto INCLUDE3; END_CHECK_NUM: if ( ++i <= n ) goto CHECK_NUM; cout << endl; }
|
n = int(input())
i = 1
while i <= n:
if i % 3 == 0:
print(' {}'.format(i))
continue
elif i % 10 == 3:
print(' {}'.format(i))
continue
|
s196356666
|
Accepted
| 40 | 7,964 | 227 |
n = int(input())
i = 1
for i in range(1, n + 1):
if i % 3 == 0:
print(' {0}'.format(i), end = '')
else:
st = str(i)
for x in range(0, len(st)):
if st[x] == '3':
print(' {0}'.format(i), end = '')
break
print()
|
s172833995
|
p03090
|
u648212584
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,064 | 830 |
You are given an integer N. Build an undirected graph with N vertices with indices 1 to N that satisfies the following two conditions: * The graph is simple and connected. * There exists an integer S such that, for every vertex, the sum of the indices of the vertices adjacent to that vertex is S. It can be proved that at least one such graph exists under the constraints of this problem.
|
import sys
input = sys.stdin.readline
def main():
N = int(input())
if N == 3:
print(1,3)
print(2,3)
elif N == 4:
print(1,2)
print(1,3)
print(4,2)
print(4,3)
else:
if N%2 == 0:
for i in range(N//2-1):
print(i+1,i+2)
print(i+1,N-i-1)
print(N-i,i+2)
print(N-i,N-i-1)
for i in range(2):
print(N//2+i,1)
print(N//2+i,N)
else:
for i in range(N//2-1):
print(i+1,i+2)
print(i+1,N-i-2)
print(N-i-1,i+2)
print(N-i-1,N-i-2)
print(N//2,N)
print(N//2+1,N)
print(N,1)
print(N,N-1)
if __name__ == "__main__":
main()
|
s731674467
|
Accepted
| 17 | 3,188 | 912 |
import sys
input = sys.stdin.readline
def main():
N = int(input())
if N == 3:
print(2)
print(1,3)
print(2,3)
elif N == 4:
print(4)
print(1,2)
print(1,3)
print(4,2)
print(4,3)
else:
if N%2 == 0:
print(N*2)
for i in range(N//2-1):
print(i+1,i+2)
print(i+1,N-i-1)
print(N-i,i+2)
print(N-i,N-i-1)
for i in range(2):
print(N//2+i,1)
print(N//2+i,N)
else:
print(N*2-2)
for i in range(N//2-1):
print(i+1,i+2)
print(i+1,N-i-2)
print(N-i-1,i+2)
print(N-i-1,N-i-2)
print(N//2,N)
print(N//2+1,N)
print(N,1)
print(N,N-1)
if __name__ == "__main__":
main()
|
s535000081
|
p02606
|
u326408598
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,156 | 158 |
How many multiples of d are there among the integers between L and R (inclusive)?
|
l, r,d = map(int,input().split())
r-=1
count=0
if r<d:
print(count)
else:
if l<d:
l=d
count+=1
count+=((r-l)//d)
print(count)
|
s036557526
|
Accepted
| 25 | 9,176 | 243 |
l, r,d = map(int,input().split())
count=0
if r<d:
print(count)
else:
if l<d:
l=d
count+=1
elif l%d==0:
count+=1
elif l%d!=0:
l += d-(l%d)
count+=1
count+=((r-l)//d)
print(count)
|
s520226125
|
p03394
|
u211160392
| 2,000 | 262,144 |
Wrong Answer
| 44 | 3,424 | 209 |
Nagase is a top student in high school. One day, she's analyzing some properties of special sets of positive integers. She thinks that a set S = \\{a_{1}, a_{2}, ..., a_{N}\\} of **distinct** positive integers is called **special** if for all 1 \leq i \leq N, the gcd (greatest common divisor) of a_{i} and the sum of the remaining elements of S is **not** 1. Nagase wants to find a **special** set of size N. However, this task is too easy, so she decided to ramp up the difficulty. Nagase challenges you to find a **special** set of size N such that the gcd of all elements are 1 and the elements of the set does not exceed 30000.
|
N = int(input())
a,b = 0,0
for i in range(1,5001):
a = 3*i
b = N-a
if b <= 5000 and b%2 == 0:
break
for i in range(a):
print(2*i+2,end=" ")
for i in range(b):
print(6*i+3,end=" ")
|
s793571956
|
Accepted
| 45 | 3,424 | 442 |
N = int(input())
if N == 3:
print(2,5,63)
exit()
if N == 4:
print(2,5,20,63)
exit()
if N == 6:
print(2,4,6,3,9,24)
exit()
if N == 19999:
for i in range(1,30000):
if i%2 == 0 or i % 3 == 0:
print(i,end=" ")
exit()
for a in range(3,15001,3):
b = N-a
if 1 < b <= 5000 and b%2 == 0:
break
for i in range(a):
print(2*i+2,end=" ")
for i in range(b):
print(6*i+3,end=" ")
|
s563133568
|
p03681
|
u536034761
| 2,000 | 262,144 |
Wrong Answer
| 523 | 10,292 | 291 |
Snuke has N dogs and M monkeys. He wants them to line up in a row. As a Japanese saying goes, these dogs and monkeys are on bad terms. _("ken'en no naka", literally "the relationship of dogs and monkeys", means a relationship of mutual hatred.)_ Snuke is trying to reconsile them, by arranging the animals so that there are neither two adjacent dogs nor two adjacent monkeys. How many such arrangements there are? Find the count modulo 10^9+7 (since animals cannot understand numbers larger than that). Here, dogs and monkeys are both distinguishable. Also, two arrangements that result from reversing each other are distinguished.
|
from math import factorial
def combinations_count(n, r):
return factorial(n) // (factorial(n - r) * factorial(r))
N, M = map(int, input().split())
mod = 10**9 + 7
print(((factorial(N) % mod) * (combinations_count(N + 1, M) % mod) * (factorial(M) % mod)) % mod if abs(N - M) <= 2 else 0)
|
s274673246
|
Accepted
| 275 | 10,048 | 258 |
from math import factorial
def count(n, r):
if n == r:
return 2
else:
return 1
N, M = map(int, input().split())
mod = 10**9 + 7
print(((factorial(N) % mod) * (count(N, M) % mod) * (factorial(M) % mod)) % mod if abs(N - M) < 2 else 0)
|
s734068767
|
p02260
|
u002280517
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 306 |
Write a program of the Selection Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: SelectionSort(A) 1 for i = 0 to A.length-1 2 mini = i 3 for j = i to A.length-1 4 if A[j] < A[mini] 5 mini = j 6 swap A[i] and A[mini] Note that, indices for array elements are based on 0-origin. Your program should also print the number of swap operations defined in line 6 of the pseudocode in the case where i ≠mini.
|
n = int(input())
a = list(map(int,input().split()))
count = 0
for i in range(n):
minij =i
for j in range(i,n):
if a[j] < a[minij]:
minij = j
a[i],a[minij] = a[minij],a[i]
count+=1
for i in range(n):
print("{0} ".format(a[i]),end="")
print()
print(count)
|
s387306320
|
Accepted
| 20 | 5,600 | 292 |
n = int(input())
a = list(map(int,input().split()))
count = 0
for i in range(n):
minij =i
for j in range(i,n):
if a[j] < a[minij]:
minij = j
if a[i] > a[minij]:
count+=1
a[i],a[minij] = a[minij],a[i]
print(' '.join(map(str,a)))
print(count)
|
s715262563
|
p02694
|
u727787724
| 2,000 | 1,048,576 |
Wrong Answer
| 22 | 9,044 | 65 |
Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank. The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.) Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?
|
n=int(input())
z=100
i=0
while z>n:
z=z**(1.01)
i+=1
print(i)
|
s234127459
|
Accepted
| 25 | 9,220 | 121 |
# coding: utf-8
# Your code here!
n=int(input())
z=100
i=0
while z<n:
z=z*(1.01)
z=z//1
i+=1
#print(z)
print(i)
|
s900426247
|
p02613
|
u011202375
| 2,000 | 1,048,576 |
Wrong Answer
| 155 | 17,408 | 309 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
S=input()
S=int(S)
a = [input() for i in range(S)]
AC=0
WA=0
TLE=0
RE=0
print(a)
for a in a:
if a=="AC":
AC+=1
if a=="WA":
WA+=1
if a=="TLE":
TLE+=1
if a=="RE":
RE+=1
print("AC x "+str(AC) )
print("WA x "+str(WA))
print("TLE x "+str(TLE))
print("RE x "+str(RE))
|
s603502803
|
Accepted
| 156 | 15,952 | 301 |
S=input()
S=int(S)
a = [input() for i in range(S)]
AC=0
WA=0
TLE=0
RE=0
for a in a:
if a=="AC":
AC+=1
if a=="WA":
WA+=1
if a=="TLE":
TLE+=1
if a=="RE":
RE+=1
print("AC x "+str(AC) )
print("WA x "+str(WA))
print("TLE x "+str(TLE))
print("RE x "+str(RE))
|
s160951245
|
p03523
|
u760961723
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 294 |
You are given a string S. Takahashi can insert the character `A` at any position in this string any number of times. Can he change S into `AKIHABARA`?
|
S = input()
ans_list = ["AKIHABARA",\
"KIHABARA","AKIHBARA","AKIHABRA","AKIHABAR",\
"KIHBARA","KIHABRA","KIHABAR","AKIHBRA","AKIHBAR","AKIHABR",\
"AKIHBR","KIHABR","KIHBAR","KIHBRA"\
"KIHBR"]
if S in ans_list:
print("YES")
else:
print("NO")
|
s427576019
|
Accepted
| 17 | 2,940 | 297 |
S = input()
ans_list = ["AKIHABARA",\
"KIHABARA","AKIHBARA","AKIHABRA","AKIHABAR",\
"KIHBARA","KIHABRA","KIHABAR","AKIHBRA","AKIHBAR","AKIHABR",\
"AKIHBR","KIHABR","KIHBAR","KIHBRA",\
"KIHBR"]
if S in ans_list:
print("YES")
else:
print("NO")
|
s048162199
|
p03997
|
u864900001
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 72 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
#45
a = int(input())
ans = (a+ int(input()))/2
print(ans * int(input()))
|
s753003739
|
Accepted
| 17 | 2,940 | 77 |
#45
a = int(input())
ans = (a+ int(input()))/2
print(int(ans * int(input())))
|
s190744786
|
p04012
|
u118019047
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 70 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
q = input()
if len(q) % 2 == 0:
print("YES")
else:
print("NO")
|
s270995786
|
Accepted
| 17 | 2,940 | 104 |
w = input()
for i in w:
if w.count(i)%2!=0:
print('No')
break
else:
print('Yes')
|
s427301885
|
p02606
|
u647835149
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,160 | 107 |
How many multiples of d are there among the integers between L and R (inclusive)?
|
list = list(map(int, input().split()))
sho1 = list[1] // list[2]
sho2 = list[0] // list[2]
print(sho2-sho1)
|
s958682140
|
Accepted
| 34 | 8,996 | 191 |
list = list(map(int,input().split()))
large = list[1] // list[2]
small = list[0] // list[2]
amari = list[0] % list[2]
if amari == 0:
print(large - small + 1)
else:
print(large - small)
|
s764579753
|
p02613
|
u262039080
| 2,000 | 1,048,576 |
Wrong Answer
| 158 | 16,164 | 307 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N=int(input())
S=[0]*(N+1)
T=[0]*4
for i in range(N):
S[i]=input()
for i in range(N):
if S[i]=="AC":
T[0]+=1
elif S[i]=="WA":
T[1]+=1
elif S[i]=="TLE":
T[2]+=1
else:
T[3]+=1
print("AC ×",T[0])
print("WA ×",T[1])
print("TLE ×",T[2])
print("RE ×",T[3])
|
s528410202
|
Accepted
| 159 | 16,044 | 303 |
N=int(input())
S=[0]*(N+1)
T=[0]*4
for i in range(N):
S[i]=input()
for i in range(N):
if S[i]=="AC":
T[0]+=1
elif S[i]=="WA":
T[1]+=1
elif S[i]=="TLE":
T[2]+=1
else:
T[3]+=1
print("AC x",T[0])
print("WA x",T[1])
print("TLE x",T[2])
print("RE x",T[3])
|
s949242029
|
p04029
|
u720417458
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 70 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N = int(input())
a = 0
for i in range(N):
a += i + 1
print(a)
|
s494305106
|
Accepted
| 17 | 2,940 | 40 |
N = int(input())
a = N*(N+1)//2
print(a)
|
s629046442
|
p03698
|
u626468554
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 286 |
You are given a string S consisting of lowercase English letters. Determine whether all the characters in S are different.
|
#n = int(input())
#n,k = map(int,input().split())
#x = list(map(int,input().split()))
#62
s = list(input().split())
jg = {}
for i in range(len(s)):
if s[i] in jg:
print("no")
break
else:
jg[s[i]] = 1
if i == len(s):
print("yes")
|
s590499844
|
Accepted
| 17 | 2,940 | 281 |
#n = int(input())
#n,k = map(int,input().split())
#x = list(map(int,input().split()))
#62
s = list(input())
jg = {}
for i in range(len(s)):
if s[i] in jg:
print("no")
break
else:
jg[s[i]] = 1
if i == len(s)-1:
print("yes")
|
s995147841
|
p02613
|
u589361760
| 2,000 | 1,048,576 |
Wrong Answer
| 164 | 16,080 | 334 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n = int(input())
s = [input() for i in range(n)]
i = 0
a = 0
t = 0
w = 0
r = 0
for i in range(n):
if s[i] == "AC":
a += 1
i += 1
elif s[i] == "TLE":
t += 1
i += 1
elif s[i] == "WA":
w += 1
i += 1
else:
r += 1
i += 1
print(. format(a, w, t, r))
|
s070798359
|
Accepted
| 167 | 16,316 | 331 |
n = int(input())
s = [input() for i in range(n)]
i = 0
a = 0
t = 0
w = 0
r = 0
for i in range(n):
if s[i] == "AC":
a += 1
i += 1
elif s[i] == "TLE":
t += 1
i += 1
elif s[i] == "WA":
w += 1
i += 1
else:
r += 1
i += 1
print("""AC x {0}
WA x {1}
TLE x {2}
RE x {3}""". format(a, w, t, r))
|
s465051515
|
p03025
|
u268793453
| 2,000 | 1,048,576 |
Wrong Answer
| 302 | 27,664 | 876 |
Takahashi and Aoki will play a game. They will repeatedly play it until one of them have N wins in total. When they play the game once, Takahashi wins with probability A %, Aoki wins with probability B %, and the game ends in a draw (that is, nobody wins) with probability C %. Find the expected number of games that will be played, and print it as follows. We can represent the expected value as P/Q with coprime integers P and Q. Print the integer R between 0 and 10^9+6 (inclusive) such that R \times Q \equiv P\pmod {10^9+7}. (Such an integer R always uniquely exists under the constraints of this problem.)
|
n, a, b, c = [int(i) for i in input().split()]
p = 10 ** 9 + 7
ans = 0
def fact(n, p=10**9 + 7):
f = [1]
for i in range(1, n+1):
f.append(f[-1]*i%p)
return f
def invfact(n, f, p=10**9 + 7):
inv = [pow(f[n], p-2, p)]
for i in range(n, 0, -1):
inv.append(inv[-1]*i%p)
return inv[::-1]
f = fact(2 * n)
invf = invfact(2 * n, f)
def comb(a, b):
if a < b:
return 0
if a < 0 or b < 0:
return 0
return f[a] * invf[b] * invf[a-b] % p
pow_a = [1]
pow_b = [1]
inv_100 = pow(100, p - 2, p)
a = a * inv_100 % p
b = b * inv_100 % p
c = c * inv_100 % p
for i in range(n):
pow_a.append(pow_a[-1] * a % p)
pow_b.append(pow_b[-1] * b % p)
for m in range(n, 2 * n):
ans += comb(m - 1, n - 1) * (pow_a[n] * pow_b[m - n] + pow_a[m - n] * pow_b[n]) * m
ans %= p
ans *= pow(1 - c, p - 2, p)
print(ans % p)
|
s442257842
|
Accepted
| 301 | 27,944 | 945 |
n, a, b, c = [int(i) for i in input().split()]
p = 10 ** 9 + 7
ans = 0
def fact(n, p=10**9 + 7):
f = [1]
for i in range(1, n+1):
f.append(f[-1]*i%p)
return f
def invfact(n, f, p=10**9 + 7):
inv = [pow(f[n], p-2, p)]
for i in range(n, 0, -1):
inv.append(inv[-1]*i%p)
return inv[::-1]
f = fact(2 * n)
invf = invfact(2 * n, f)
def comb(a, b):
if a < b:
return 0
if a < 0 or b < 0:
return 0
return f[a] * invf[b] * invf[a-b] % p
pow_a = [1]
pow_b = [1]
inv_100 = pow(100, p - 2, p)
a = a * inv_100 % p
b = b * inv_100 % p
c = c * inv_100 % p
inv_ab = pow(a + b, p - 2, p)
a = a * inv_ab % p
b = b * inv_ab % p
for i in range(n):
pow_a.append(pow_a[-1] * a % p)
pow_b.append(pow_b[-1] * b % p)
for m in range(n, 2 * n):
ans += comb(m - 1, n - 1) * (pow_a[n] * pow_b[m - n] + pow_a[m - n] * pow_b[n]) * m
ans %= p
ans *= pow(1 - c, p - 2, p)
print(ans % p)
|
s931351707
|
p03160
|
u295242763
| 2,000 | 1,048,576 |
Wrong Answer
| 24 | 9,072 | 218 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
def aFrog(N,h):
dp = [0 for i in range(10)]
dp[0] = 0
dp[1] = abs(h[1]-h[0])
print(dp)
for i in range(2,N):
dp[i] = min(dp[i-1]+abs(h[i]-h[i-1]), dp[i-2]+abs(h[i]-h[i-2]))
return dp[N-1]
|
s347338572
|
Accepted
| 126 | 20,600 | 262 |
N = int(input())
h = list(map(int, input().split()))
inf = 10**9
dp = [0] + [inf]*N
for i in range(1,N):
if i == 1:
dp[1] = abs(h[1]-h[0])
else:
dp[i] = min(dp[i-1]+abs(h[i]-h[i-1]), dp[i-2]+abs(h[i]-h[i-2]))
print(dp[N-1])
|
s582471397
|
p03712
|
u606523772
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,060 | 263 |
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
H, W = map(int, input().split())
A_list = []
for i in range(H):
A = input()
A_list.append(A)
str1 = ["*" for _ in range(W+2)]
str1 = "".join(str1)
for i in range(H+2):
if i==0 or i==H+1:
print(str1)
else:
print("*"+A_list[i-1]+"*")
|
s719277059
|
Accepted
| 17 | 3,060 | 263 |
H, W = map(int, input().split())
A_list = []
for i in range(H):
A = input()
A_list.append(A)
str1 = ["#" for _ in range(W+2)]
str1 = "".join(str1)
for i in range(H+2):
if i==0 or i==H+1:
print(str1)
else:
print("#"+A_list[i-1]+"#")
|
s268337793
|
p02618
|
u285443936
| 2,000 | 1,048,576 |
Wrong Answer
| 82 | 9,412 | 694 |
AtCoder currently hosts three types of contests: ABC, ARC, and AGC. As the number of users has grown, in order to meet the needs of more users, AtCoder has decided to increase the number of contests to 26 types, from AAC to AZC. For convenience, we number these 26 types as type 1 through type 26. AtCoder wants to schedule contests for D days so that user satisfaction is as high as possible. For every day, AtCoder will hold exactly one contest, and each contest will end on that day. The satisfaction is calculated as follows. * The satisfaction at the beginning of day 1 is 0. Satisfaction can be negative. * Holding contests increases satisfaction. The amount of increase will vary depending on a variety of factors. Specifically, we know in advance that holding a contest of type i on day d will increase the satisfaction by s_{d,i}. * If a particular type of contest is not held for a while, the satisfaction decreases. Each contest type i has an integer c_i, and at the end of each day d=1,2,...,D, the satisfaction decreases as follows. Let \mathrm{last}(d,i) be the last day before day d (including d) on which a contest of type i was held. If contests of type i have never been held yet, we define \mathrm{last}(d,i)=0. At the end of day d, the satisfaction decreases by \sum _{i=1}^{26}c_i \times (d-\mathrm{last}(d,i)). Please schedule contests on behalf of AtCoder. If the satisfaction at the end of day D is S, you will get a score of \max(10^6 + S, 0). There are 50 test cases, and the score of a submission is the total scores for each test case. You can make submissions multiple times, and the highest score among your submissions will be your score.
|
D = int(input())
c = list(map(int,input().split()))
s = list(list(map(int,input().split())) for i in range(D))
t = []
score = 0
last = [0]*26
#v = [0]*D
def calc(d):
score = 0
t = 0
contest = 0
for i in range(26):
temp = 0
temp += s[d][i]
for j in range(26):
temp -= c[j]*(d+1-last[j])
print(score,temp)
if score > temp:
continue
else:
score = temp
contest = i
return contest
for d in range(D):
i = calc(d)
t.append(i)
last[i] = d+1
for i in range(D):
print(t[i])
|
s867632010
|
Accepted
| 80 | 9,304 | 670 |
D = int(input())
c = list(map(int,input().split()))
s = list(list(map(int,input().split())) for i in range(D))
t = []
score = 0
last = [0]*26
#v = [0]*D
def calc(d):
score = 0
t = 0
contest = 0
for i in range(26):
temp = 0
temp += s[d][i]
for j in range(26):
temp -= c[j]*(d+1-last[j])
if score > temp:
continue
else:
score = temp
contest = i
return contest
for d in range(D):
i = calc(d)
t.append(i)
last[i] = d+1
for i in range(D):
print(t[i]+1)
|
s960130477
|
p02264
|
u928329738
| 1,000 | 131,072 |
Wrong Answer
| 50 | 6,984 | 481 |
_n_ _q_ _name 1 time1_ _name 2 time2_ ... _name n timen_ In the first line the number of processes _n_ and the quantum _q_ are given separated by a single space. In the following _n_ lines, names and times for the _n_ processes are given. _name i_ and _time i_ are separated by a single space.
|
import queue
procs = queue.Queue()
n,q = list(map(int,input().split()))
for i in range(n):
name,time = input().split()
procs.put((name,int(time)))
time=0
doneprocs = queue.Queue()
while True:
if procs.empty():
break
temp = procs.get()
if temp[1] < q:
time += temp[1]
doneprocs.put((temp[0],time))
else:
time+= q
procs.put((temp[0],temp[1]-q))
for i in range(n):
temp = doneprocs.get()
print(temp[0],temp[1])
|
s689789688
|
Accepted
| 2,120 | 15,172 | 482 |
import queue
procs = queue.Queue()
n,q = list(map(int,input().split()))
for i in range(n):
name,time = input().split()
procs.put((name,int(time)))
time=0
doneprocs = queue.Queue()
while True:
if procs.empty():
break
temp = procs.get()
if temp[1] <= q:
time += temp[1]
doneprocs.put((temp[0],time))
else:
time+= q
procs.put((temp[0],temp[1]-q))
for i in range(n):
temp = doneprocs.get()
print(temp[0],temp[1])
|
s806222589
|
p02612
|
u425236751
| 2,000 | 1,048,576 |
Wrong Answer
| 30 | 9,156 | 123 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
i = 1
while(True):
change = N - 1000*i
if change >= 1000:
i+=1
else:
break
print(change)
|
s153046026
|
Accepted
| 31 | 8,888 | 118 |
N = int(input())
i = 1
while(True):
change = 1000*i - N
if change < 0:
i+=1
else:
break
print(change)
|
s824918180
|
p03548
|
u998835868
| 2,000 | 262,144 |
Wrong Answer
| 49 | 3,064 | 303 |
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters. We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person. At most how many people can sit on the seat?
|
x,y,z = map(int,input().split())
aa = x
b1 = 0
b2 = 0
for a in range(aa):
x -= y
if x <= 0:
break
x -= z
if x < 0:
break
b1 += 1
x = aa
for a in range(aa):
x -= z
if x <= 0:
break
x -= y
if x < 0:
break
b2 += 1
print(b1,b2)
if b1 <= b2:
print(b2)
else:
print(b1)
|
s825837594
|
Accepted
| 33 | 3,060 | 188 |
x,y,z = map(int,input().split())
aa = x
b1 = 0
b2 = 0
for a in range(aa):
x -= z
if x < y:
break
x -= y
if x < z:
break
b2 += 1
if b1 <= b2:
print(b2)
else:
print(b1)
|
s115433554
|
p02612
|
u122495382
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 9,028 | 38 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
a = n % 1000
print(a)
|
s674878054
|
Accepted
| 26 | 9,160 | 74 |
n = int(input())
a = n % 1000
if a == 0:
print(a)
else:
print(1000- a)
|
s387939645
|
p03997
|
u002459665
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 74 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
a = int(input())
b = int(input())
h = int(input())
print((a + b) * h / 2)
|
s394404304
|
Accepted
| 18 | 2,940 | 91 |
a = int(input())
b = int(input())
h = int(input())
ans = (a + b) * h / 2
print(round(ans))
|
s510969683
|
p03610
|
u587213169
| 2,000 | 262,144 |
Wrong Answer
| 30 | 4,336 | 76 |
You are given a string s consisting of lowercase English letters. Extract all the characters in the odd-indexed positions and print the string obtained by concatenating them. Here, the leftmost character is assigned the index 1.
|
s=input()
N=len(s)
ans=[]
for i in range(0, N, 2):
ans+=s[i]
print(ans)
|
s495147221
|
Accepted
| 28 | 3,188 | 79 |
s=input()
N=len(s)
ans=str()
for i in range(0, N, 2):
ans+=s[i]
print(ans)
|
s955676058
|
p03095
|
u760171369
| 2,000 | 1,048,576 |
Wrong Answer
| 18 | 3,188 | 126 |
You are given a string S of length N. Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings. Here, a subsequence of a string is a concatenation of **one or more** characters from the string without changing the order.
|
N = int(input())
S = input().split()
cnt = 1
L = list(set(S))
for k in L:
cnt *= (S.count(k) + 1)
print(cnt % (10 ** 9 + 7))
|
s542095650
|
Accepted
| 22 | 3,188 | 117 |
N = input()
S = input()
L = list(set(S))
cnt = 1
for k in L:
cnt *= (S.count(k) + 1)
print((cnt - 1) % (10**9 + 7))
|
s054420119
|
p03695
|
u243699903
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 573 |
In AtCoder, a person who has participated in a contest receives a _color_ , which corresponds to the person's rating as follows: * Rating 1-399 : gray * Rating 400-799 : brown * Rating 800-1199 : green * Rating 1200-1599 : cyan * Rating 1600-1999 : blue * Rating 2000-2399 : yellow * Rating 2400-2799 : orange * Rating 2800-3199 : red Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not. Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i. Find the minimum and maximum possible numbers of different colors of the users.
|
n = int(input())
color = [0] * 8
pro = 0
for ai in map(int, input().split()):
print(ai)
if 1 <= ai <= 399:
color[0] = 1
elif 400 <= ai <= 799:
color[1] = 1
elif 800 <= ai <= 1199:
color[2] = 1
elif 1200 <= ai <= 1599:
color[3] = 1
elif 1600 <= ai <= 1999:
color[4] = 1
elif 2000 <= ai <= 2399:
color[5] = 1
elif 2400 <= ai <= 2799:
color[6] = 1
elif 2800 <= ai <= 3199:
color[7] = 1
else:
pro += 1
colSum = sum(color)
print(colSum, colSum+min(8-colSum, pro))
|
s518233412
|
Accepted
| 17 | 3,064 | 602 |
n = int(input())
color = [0] * 8
pro = 0
for ai in map(int, input().split()):
if 1 <= ai <= 399:
color[0] = 1
elif 400 <= ai <= 799:
color[1] = 1
elif 800 <= ai <= 1199:
color[2] = 1
elif 1200 <= ai <= 1599:
color[3] = 1
elif 1600 <= ai <= 1999:
color[4] = 1
elif 2000 <= ai <= 2399:
color[5] = 1
elif 2400 <= ai <= 2799:
color[6] = 1
elif 2800 <= ai <= 3199:
color[7] = 1
else:
pro += 1
colSum = sum(color)
if colSum == 0 and pro > 0:
print(1, pro)
else:
print(colSum, colSum + pro)
|
s043094953
|
p03251
|
u793982420
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,064 | 566 |
Our world is one-dimensional, and ruled by two empires called Empire A and Empire B. The capital of Empire A is located at coordinate X, and that of Empire B is located at coordinate Y. One day, Empire A becomes inclined to put the cities at coordinates x_1, x_2, ..., x_N under its control, and Empire B becomes inclined to put the cities at coordinates y_1, y_2, ..., y_M under its control. If there exists an integer Z that satisfies all of the following three conditions, they will come to an agreement, but otherwise war will break out. * X < Z \leq Y * x_1, x_2, ..., x_N < Z * y_1, y_2, ..., y_M \geq Z Determine if war will break out.
|
# -*- coding: utf-8 -*-
"""
Created on Sun Sep 23 21:12:01 2018
@author: masato
"""
N, M, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
x.sort()
y.sort()
xmin = x[0]
ymin = y[0]
xmax = x[-1]
ymax = y[-1]
z = max(xmax,ymin)
iswar = 0
for i in range(len(x)):
for j in range(len(y)):
if x[i] > z:
iswar = 1
break
elif y[j] <= z:
iswar = 1
break
if iswar == 1:
break
if iswar == 1:
print("War")
else:
print("No War")
|
s424445408
|
Accepted
| 17 | 3,064 | 542 |
# -*- coding: utf-8 -*-
"""
Created on Sun Sep 23 21:12:01 2018
@author: masato
"""
N, M, X, Y = map(int, input().split())
x = list(map(int, input().split()))
y = list(map(int, input().split()))
x.sort()
y.sort()
xmin = x[0]
ymin = y[0]
xmax = x[-1]
ymax = y[-1]
z = max(xmax,ymin)
iswar = 0
for i in range(len(x)):
if min(ymin,Y) <= x[i]:
iswar = 1
break
for j in range(len(y)):
if max(xmax,X) >= y[j]:
iswar = 1
break
if iswar == 1:
print("War")
else:
print("No War")
|
s677615816
|
p03477
|
u808427016
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 308 |
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
import sys
def solve1(s):
n = len(s)
i = 0
while i <= n // 2:
if s[n // 2] != s[n // 2 - i]:
break
if s[(n - 1) // 2] != s[(n - 1) // 2 + i]:
break
i += 1
return (n - 1) // 2 + i
line = sys.stdin.readline().strip()
print(solve1(list(line)))
|
s680657979
|
Accepted
| 17 | 2,940 | 197 |
import sys
line = sys.stdin.readline().strip()
a, b, c, d = [int(x) for x in line.split()]
if a + b == c + d:
print("Balanced")
elif a + b > c + d:
print("Left")
else:
print("Right")
|
s839835648
|
p04011
|
u268792407
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 115 |
There is a hotel with the following accommodation fee: * X yen (the currency of Japan) per night, for the first K nights * Y yen per night, for the (K+1)-th and subsequent nights Tak is staying at this hotel for N consecutive nights. Find his total accommodation fee.
|
n=int(input())
k=int(input())
x=int(input())
y=int(input())
if n<=k:
print(x * n)
else:
print(k * n +(n-k) * y)
|
s002324248
|
Accepted
| 17 | 2,940 | 115 |
n=int(input())
k=int(input())
x=int(input())
y=int(input())
if n<=k:
print(x * n)
else:
print(k * x +(n-k) * y)
|
s344037746
|
p03644
|
u819135704
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 257 |
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
n = int(input())
count = 0
list_count = []
max_number = 0
for i in range(n):
a = i + 1
while a % 2 == 0:
count += 1
a /= 2
list_count.append(count)
if max(list_count) == count:
max_number = i + 1
print(max_number)
|
s929242621
|
Accepted
| 18 | 2,940 | 271 |
n = int(input())
count = 0
list_count = []
max_number = 0
for i in range(n):
a = i + 1
while a % 2 == 0:
count += 1
a /= 2
list_count.append(count)
if max(list_count) == count:
max_number = i + 1
count = 0
print(max_number)
|
s944863323
|
p02613
|
u737842024
| 2,000 | 1,048,576 |
Wrong Answer
| 143 | 9,108 | 266 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
N=int(input())
AC=0
WA=0
TLE=0
RE=0
for i in range(N):
c=input()
if(c=='AC'):
AC+=1
elif(c=='WA'):
WA+=1
elif(c=='TLE'):
TLE+=1
else:
RE+=1
print("AC ×",AC)
print("WA ×",WA)
print("TLE ×",TLE)
print("RE ×",RE)
|
s394012582
|
Accepted
| 146 | 9,208 | 262 |
N=int(input())
AC=0
WA=0
TLE=0
RE=0
for i in range(N):
c=input()
if(c=='AC'):
AC+=1
elif(c=='WA'):
WA+=1
elif(c=='TLE'):
TLE+=1
else:
RE+=1
print("AC x",AC)
print("WA x",WA)
print("TLE x",TLE)
print("RE x",RE)
|
s253230963
|
p03455
|
u189636973
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 122 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
import sys
inf = sys.maxsize
a, b = map(int, input().split())
if(a%2==0 and b%2==0):
print("Even")
else:
print("Odd")
|
s143633089
|
Accepted
| 17 | 2,940 | 86 |
a, b = map(int, input().split())
if(a*b % 2 == 0):
print("Even")
else:
print("Odd")
|
s457081729
|
p03544
|
u598229387
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 141 |
It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers. You are given an integer N. Find the N-th Lucas number. Here, the i-th Lucas number L_i is defined as follows: * L_0=2 * L_1=1 * L_i=L_{i-1}+L_{i-2} (i≥2)
|
n = int(input())
a,b,c = 2,1,3
for i in range(n-3):
a = b+c
b = a+c
c = a+b
print(a if n>=4 else b if n==2 else c if n==3 else a)
|
s818096644
|
Accepted
| 17 | 2,940 | 70 |
n = int(input())
a,b = 2,1
for i in range(n):
a,b = b,a+b
print(a)
|
s021977557
|
p04012
|
u740284863
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 162 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
w = str(input())
count = 1
for i in range(len(w)):
s = w.count(w[i])
if s % 2 != 0:
print("No")
count -=1
if count == 0:
print("Yes")
|
s081054844
|
Accepted
| 29 | 2,940 | 176 |
w = str(input())
count = 1
for i in range(len(w)):
s = w.count(w[i])
if s % 2 != 0:
print("No")
count -=1
break
if count == 1:
print("Yes")
|
s410446315
|
p03386
|
u600261652
| 2,000 | 262,144 |
Wrong Answer
| 2,141 | 619,156 | 454 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
def resolve():
A, B, K = map(int, input().split())
count = [A]
ans = []
for _ in range(B-A):
count.append(A+1)
A += 1
if K > B-A-1:
K = B-A+1
for i in range(K):
if count[i] <= B:
ans.append(count[i])
for j in range(K):
if count[(j+1)*-1] >= count[K-1]:
ans.append(count[j*-1])
an = sorted(set(ans))
for i in range(len(an)):
print(an[i])
resolve()
|
s975429065
|
Accepted
| 17 | 3,060 | 252 |
def resolve():
A, B, K = map(int, input().split())
if 2*K > B-A+1:
for _ in range(B-A+1):
print(A+_)
else:
for i in range(K):
print(A+i)
for j in range(K):
print(B-K+j+1)
resolve()
|
s839698926
|
p03080
|
u765865533
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 3,060 | 137 |
There are N people numbered 1 to N. Each person wears a red hat or a blue hat. You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`. Determine if there are more people wearing a red hat than people wearing a blue hat.
|
N = int(input())
s = input()
RBlist = list(s)
print(type(N))
r=RBlist.count('R')
b=N-r
if b<r:
print("Yes")
else:
print("No")
|
s622898312
|
Accepted
| 17 | 2,940 | 138 |
N = int(input())
s = input()
RBlist = list(s)
#print(type(N))
r=RBlist.count('R')
b=N-r
if b<r:
print("Yes")
else:
print("No")
|
s123979119
|
p03816
|
u297109012
| 2,000 | 262,144 |
Wrong Answer
| 62 | 19,324 | 303 |
Snuke has decided to play a game using cards. He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written. He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept. Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.
|
from collections import Counter
def solve(N, As):
print(N, ",", As)
counter = Counter(As)
if len(counter) % 2 == 0:
return len(counter) - 1
return len(counter)
if __name__ == "__main__":
N = int(input())
As = list(map(int, input().split(" ")))
print(solve(N, As))
|
s566385193
|
Accepted
| 54 | 18,656 | 281 |
from collections import Counter
def solve(N, As):
counter = Counter(As)
if len(counter) % 2 == 0:
return len(counter) - 1
return len(counter)
if __name__ == "__main__":
N = int(input())
As = list(map(int, input().split(" ")))
print(solve(N, As))
|
s952284011
|
p02659
|
u180528413
| 2,000 | 1,048,576 |
Wrong Answer
| 23 | 9,192 | 352 |
Compute A \times B, truncate its fractional part, and print the result as an integer.
|
import sys
import math
readline = sys.stdin.readline
readall = sys.stdin.read
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prl = lambda x: print(*x ,sep='\n')
a,b = input().split(' ')
res = int(a) * float(b)
print(math.floor(res),res)
|
s394524499
|
Accepted
| 22 | 9,256 | 677 |
import sys
import math
readline = sys.stdin.readline
readall = sys.stdin.read
ns = lambda: readline().rstrip()
ni = lambda: int(readline().rstrip())
nm = lambda: map(int, readline().split())
nl = lambda: list(map(int, readline().split()))
prl = lambda x: print(*x ,sep='\n')
a,b = input().split(' ')
ash_1 = int(a[-1])
ash_2 = int(a[-2:])
bsh_1 = int(b[-2])
bsh_2 = int(b[-1])
bsh_0 = int(b[0])
kuri = (ash_1*bsh_1*10 + ash_2*bsh_2) //100
if len(a) == 1:
res = int(a)*bsh_0 + kuri
elif len(a) == 2:
a1 = a[:-1]
res = int(a)*bsh_0 + int(a1)*bsh_1 + kuri
else:
a1 = a[:-1]
a2 = a[:-2]
res = int(a)*bsh_0 + int(a1)*bsh_1 + int(a2)*bsh_2 + kuri
print(res)
|
s421181818
|
p03591
|
u572032237
| 2,000 | 262,144 |
Wrong Answer
| 30 | 9,028 | 25 |
Ringo is giving a present to Snuke. Ringo has found out that Snuke loves _yakiniku_ (a Japanese term meaning grilled meat. _yaki_ : grilled, _niku_ : meat). He supposes that Snuke likes grilled things starting with `YAKI` in Japanese, and does not like other things. You are given a string S representing the Japanese name of Ringo's present to Snuke. Determine whether S starts with `YAKI`.
|
s = input()
print(s[:-8])
|
s889947170
|
Accepted
| 26 | 9,036 | 70 |
s = input()
if 'YAKI' in s[:4]:
print('Yes')
else:
print('No')
|
s136345019
|
p03962
|
u036190609
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 36 |
AtCoDeer the deer recently bought three paint cans. The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c. Here, the color of each paint can is represented by an integer between 1 and 100, inclusive. Since he is forgetful, he might have bought more than one paint can in the same color. Count the number of different kinds of colors of these paint cans and tell him.
|
s = input()
s = set(s)
print(len(s))
|
s418983853
|
Accepted
| 17 | 2,940 | 38 |
s = input().split()
print(len(set(s)))
|
s682905366
|
p02612
|
u303711501
| 2,000 | 1,048,576 |
Wrong Answer
| 27 | 8,980 | 96 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
N = int(input())
# X = list(input())
def p(X):
print(X)
while N >=1000:
N -= 1000
p(N)
|
s572249587
|
Accepted
| 28 | 9,040 | 108 |
N = int(input())
add = 1000 - N%1000
if add == 1000:
add = 0
else :
add = 1000 - N%1000
print(add)
|
s211844041
|
p03407
|
u220870679
| 2,000 | 262,144 |
Wrong Answer
| 19 | 2,940 | 80 |
An elementary school student Takahashi has come to a variety store. He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it? Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.
|
a,b,c = map(int, input().split())
if a + b > c:
print('No')
else:
print('Yes')
|
s883737655
|
Accepted
| 17 | 2,940 | 93 |
a, b, c =map(int, input() .split())
if 0 <= a + b - c:
print("Yes")
else:
print("No")
|
s865804904
|
p03699
|
u980783809
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 238 |
You are taking a computer-based examination. The examination consists of N questions, and the score allocated to the i-th question is s_i. Your answer to each question will be judged as either "correct" or "incorrect", and your grade will be the sum of the points allocated to questions that are answered correctly. When you finish answering the questions, your answers will be immediately judged and your grade will be displayed... if everything goes well. However, the examination system is actually flawed, and if your grade is a multiple of 10, the system displays 0 as your grade. Otherwise, your grade is displayed correctly. In this situation, what is the maximum value that can be displayed as your grade?
|
a = int(input())
list=[]
for i in range(a):
list.append(int(input()))
if not sum(list)%10==0:
print(sum(list))
else:
for i in sorted(list):
if not i%10==0:
print(sum(list)-i)
break
print(0)
|
s114250441
|
Accepted
| 18 | 3,060 | 239 |
a = int(input())
list=[]
for i in range(a):
list.append(int(input()))
if not sum(list)%10==0:
print(sum(list))
else:
for i in sorted(list):
if not i%10==0:
print(sum(list)-i)
exit()
print(0)
|
s422664197
|
p03501
|
u558764629
| 2,000 | 262,144 |
Wrong Answer
| 18 | 2,940 | 85 |
You are parking at a parking lot. You can choose from the following two fee plans: * Plan 1: The fee will be A×T yen (the currency of Japan) when you park for T hours. * Plan 2: The fee will be B yen, regardless of the duration. Find the minimum fee when you park for N hours.
|
N,A,B = map(int,input().split())
x = N * B
if x < B:
print(B)
else:
print(x)
|
s044389523
|
Accepted
| 18 | 2,940 | 85 |
N,A,B = map(int,input().split())
x = N * A
if x < B:
print(x)
else:
print(B)
|
s598479408
|
p02612
|
u349863397
| 2,000 | 1,048,576 |
Wrong Answer
| 29 | 9,000 | 28 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n=int(input())
print(n%1000)
|
s922002307
|
Accepted
| 32 | 9,508 | 1,055 |
from sys import stdin, stdout
import math,sys
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict
from os import path
import bisect as bi
import heapq
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')):
#------------------Sublime--------------------------------------#
sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
def I():return (int(input()))
def In():return(map(int,input().split()))
else:
#------------------PYPY FAst I/o--------------------------------#
def I():return (int(stdin.readline()))
def In():return(map(int,stdin.readline().split()))
def dict(a):
d={}
for x in a:
if d.get(x,-1)!=-1:
d[x]+=1
else:
d[x]=1
return d
def main():
try:
n=I()
temp=(math.ceil(n/1000)*1000)-n
print(temp)
except:
pass
M = 998244353
P = 1000000007
if __name__ == '__main__':
#for _ in range(I()):main()
for _ in range(1):main()
|
s278876585
|
p02255
|
u142825584
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,596 | 345 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
def InsertionSort(l):
for idx in range(1, len(l)):
jdx = idx - 1
while jdx >= 0 and l[idx] < l[jdx]:
l[jdx], l[idx] = l[idx], l[jdx]
idx -= 1
jdx -= 1
print(" ".join(map(str, l)))
def main():
in_list = list(map(int, input().split(' ')))
InsertionSort(in_list)
main()
|
s936183938
|
Accepted
| 20 | 5,604 | 403 |
def InsertionSort(l):
print(" ".join(map(str, l)))
for idx in range(1, len(l)):
jdx = idx - 1
while jdx >= 0 and l[idx] < l[jdx]:
l[jdx], l[idx] = l[idx], l[jdx]
idx -= 1
jdx -= 1
print(" ".join(map(str, l)))
def main():
_ = input()
in_list = list(map(int, input().split(' ')))
InsertionSort(in_list)
main()
|
s148665866
|
p02534
|
u163393400
| 2,000 | 1,048,576 |
Wrong Answer
| 25 | 9,144 | 29 |
You are given an integer K. Print the string obtained by repeating the string `ACL` K times and concatenating them. For example, if K = 3, print `ACLACLACL`.
|
n=int(input())
print('ACK'*n)
|
s267905369
|
Accepted
| 29 | 9,080 | 29 |
n=int(input())
print('ACL'*n)
|
s723361812
|
p03438
|
u787456042
| 2,000 | 262,144 |
Wrong Answer
| 24 | 5,236 | 97 |
You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N. Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal. Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions **simultaneously** : * Add 2 to a_i. * Add 1 to b_j.
|
N,*A=map(int,open(0).read().split());print("YNeos"[any(a-b-A[0]+A[N]for a,b in zip(A,A[N:]))::2])
|
s235494596
|
Accepted
| 24 | 5,236 | 126 |
N,*A=map(int,open(0).read().split());print("YNeos"[sum((b-a+1)//2for a,b in zip(A[:N],A[N:])if b>a)>sum(A[N:])-sum(A[:N])::2])
|
s039715868
|
p02608
|
u178888901
| 2,000 | 1,048,576 |
Wrong Answer
| 2,205 | 9,344 | 341 |
Let f(n) be the number of triples of integers (x,y,z) that satisfy both of the following conditions: * 1 \leq x,y,z * x^2 + y^2 + z^2 + xy + yz + zx = n Given an integer N, find each of f(1),f(2),f(3),\ldots,f(N).
|
n = int(input())
ans = [0] * n
loop = int((n/6) ** 1.5) + 1
def f(i, j, k):
return i ** 2 + j ** 2 + k ** 2 + i*j + j*k + k*i
for i in range(1, loop+1):
for j in range(1, loop+1):
for k in range(1, loop+1):
res = int(f(i, j, k))
if res <= n:
ans[res] += 1
for a in ans:
print(a)
|
s363549225
|
Accepted
| 947 | 9,504 | 347 |
n = int(input())
ans = [0] * (n + 1)
loop = int(n ** .5) + 1
def f(i, j, k):
return i ** 2 + j ** 2 + k ** 2 + i*j + j*k + k*i
for i in range(1, loop+1):
for j in range(1, loop+1):
for k in range(1, loop+1):
res = int(f(i, j, k))
if res <= n:
ans[res] += 1
for a in ans[1:]:
print(a)
|
s511830520
|
p03853
|
u740767776
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 90 |
There is an image with a height of H pixels and a width of W pixels. Each of the pixels is represented by either `.` or `*`. The character representing the pixel at the i-th row from the top and the j-th column from the left, is denoted by C_{i,j}. Extend this image vertically so that its height is doubled. That is, print a image with a height of 2H pixels and a width of W pixels where the pixel at the i-th row and j-th column is equal to C_{(i+1)/2,j} (the result of division is rounded down).
|
h, q = map(int, input().split())
s =[]
s.append(input())
for i in s:
print(i)
print(i)
|
s829457243
|
Accepted
| 18 | 3,060 | 111 |
h, q = map(int, input().split())
s =[]
for i in range(h):
s.append(input())
for i in s:
print(i)
print(i)
|
s361504596
|
p03673
|
u750651325
| 2,000 | 262,144 |
Wrong Answer
| 125 | 30,892 | 362 |
You are given an integer sequence of length n, a_1, ..., a_n. Let us consider performing the following n operations on an empty sequence b. The i-th operation is as follows: 1. Append a_i to the end of b. 2. Reverse the order of the elements in b. Find the sequence b obtained after these n operations.
|
n = int(input())
a = list(map(int, input().split()))
odd = []
even = []
ans = []
for i in range(n):
if i % 2 == 0:
odd.append(a[i])
else:
even.append(a[i])
if n == 1:
print(a[0])
exit()
else:
if n % 2 == 0:
even.reverse()
ans = even + odd
else:
odd.reverse()
ans = odd + even
print(ans)
|
s219339396
|
Accepted
| 141 | 37,144 | 381 |
n = int(input())
a = list(map(int, input().split()))
odd = []
even = []
ans = []
for i in range(n):
if i % 2 == 0:
odd.append(a[i])
else:
even.append(a[i])
if n == 1:
print(a[0])
exit()
else:
if n % 2 == 0:
even.reverse()
ans = even + odd
else:
odd.reverse()
ans = odd + even
print(" ".join(map(str,ans)))
|
s318467865
|
p02646
|
u155757809
| 2,000 | 1,048,576 |
Wrong Answer
| 22 | 9,176 | 173 |
Two children are playing tag on a number line. (In the game of tag, the child called "it" tries to catch the other child.) The child who is "it" is now at coordinate A, and he can travel the distance of V per second. The other child is now at coordinate B, and she can travel the distance of W per second. He can catch her when his coordinate is the same as hers. Determine whether he can catch her within T seconds (including exactly T seconds later). We assume that both children move optimally.
|
a, v = map(int, input().split())
b, w = map(int, input().split())
t = int(input())
oni = a + v * t
nige = b + w * t
if oni >= nige:
print('Yes')
else:
print('No')
|
s269654648
|
Accepted
| 22 | 9,180 | 200 |
a, v = map(int, input().split())
b, w = map(int, input().split())
t = int(input())
if a > b:
a, b = b, a
oni = a + v * t
nige = b + w * t
if oni >= nige:
print('YES')
else:
print('NO')
|
s410930784
|
p03779
|
u488884575
| 2,000 | 262,144 |
Wrong Answer
| 2,104 | 22,556 | 95 |
There is a kangaroo at coordinate 0 on an infinite number line that runs from left to right, at time 0. During the period between time i-1 and time i, the kangaroo can either stay at his position, or perform a jump of length exactly i to the left or to the right. That is, if his coordinate at time i-1 is x, he can be at coordinate x-i, x or x+i at time i. The kangaroo's nest is at coordinate X, and he wants to travel to coordinate X as fast as possible. Find the earliest possible time to reach coordinate X.
|
x = int(input())
cnt = 0
for i in range(1,x+1):
cnt += i
if cnt >= x:
print(i)
|
s754756820
|
Accepted
| 25 | 3,064 | 109 |
x = int(input())
cnt = 0
for i in range(1,x+1):
cnt += i
if cnt >= x:
print(i)
break
|
s474353481
|
p03361
|
u830054172
| 2,000 | 262,144 |
Wrong Answer
| 24 | 3,064 | 533 |
We have a canvas divided into a grid with H rows and W columns. The square at the i-th row from the top and the j-th column from the left is represented as (i, j). Initially, all the squares are white. square1001 wants to draw a picture with black paint. His specific objective is to make Square (i, j) black when s_{i, j}= `#`, and to make Square (i, j) white when s_{i, j}= `.`. However, since he is not a good painter, he can only choose two squares that are horizontally or vertically adjacent and paint those squares black, for some number of times (possibly zero). He may choose squares that are already painted black, in which case the color of those squares remain black. Determine if square1001 can achieve his objective.
|
h, w = list(map(int, input().split()))
s = [input() for _ in range(h)]
move = [(-1, 0), (1, 0), (0, -1), (0, 1)]
flag = 1
for i in range(h):
for j in range(w):
hantei = []
if s[i][j] == "#":
for dx, dy in move:
if i+dx<0 or j+dy<0 or i+dx>h-1 or j+dy>w-1:
continue
else:
hantei.append(s[i+dx][j+dy])
if "#" not in hantei:
flag = 0
break
if flag:
print("YES")
else:
print("NO")
|
s142483677
|
Accepted
| 23 | 3,064 | 533 |
h, w = list(map(int, input().split()))
s = [input() for _ in range(h)]
move = [(-1, 0), (1, 0), (0, -1), (0, 1)]
flag = 1
for i in range(h):
for j in range(w):
hantei = []
if s[i][j] == "#":
for dx, dy in move:
if i+dx<0 or j+dy<0 or i+dx>h-1 or j+dy>w-1:
continue
else:
hantei.append(s[i+dx][j+dy])
if "#" not in hantei:
flag = 0
break
if flag:
print("Yes")
else:
print("No")
|
s171100987
|
p02393
|
u682153677
| 1,000 | 131,072 |
Wrong Answer
| 30 | 5,572 | 79 |
Write a program which reads three integers, and prints them in ascending order.
|
# -*- coding: utf-8 -*-
a = list(map(int, input().split()))
a.sort()
print(a)
|
s289781282
|
Accepted
| 20 | 5,592 | 116 |
# -*- coding: utf-8 -*-
a = list(map(int, input().split()))
a.sort()
print('{0} {1} {2}'.format(a[0], a[1], a[2]))
|
s295097679
|
p04043
|
u961357158
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 120 |
Iroha loves _Haiku_. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order. To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.
|
# -*- coding: utf-8 -*-
A = list(map(int,input().split()))
A.sort()
if A == [5,5,7]:
print("Yes")
else:
print("No")
|
s680319467
|
Accepted
| 18 | 2,940 | 120 |
# -*- coding: utf-8 -*-
A = list(map(int,input().split()))
A.sort()
if A == [5,5,7]:
print("YES")
else:
print("NO")
|
s614315856
|
p03644
|
u513081876
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 105 |
Takahashi loves numbers divisible by 2. You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique. Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder. For example, * 6 can be divided by 2 once: 6 -> 3. * 8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1. * 3 can be divided by 2 zero times.
|
N = int(input())
ans = 0
for i in range(1, N +1):
if i % 2 == 0:
ans = max(ans, i)
print(ans)
|
s555790198
|
Accepted
| 17 | 2,940 | 93 |
N = int(input())
for i in range(1, 8):
if 2**i > N:
print(2**(i-1))
break
|
s582779858
|
p03693
|
u341736906
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 94 |
AtCoDeer has three cards, one red, one green and one blue. An integer between 1 and 9 (inclusive) is written on each card: r on the red card, g on the green card and b on the blue card. We will arrange the cards in the order red, green and blue from left to right, and read them as a three-digit integer. Is this integer a multiple of 4?
|
r,g,b = map(int,input().split())
if (100*r+10*g+b) % 4 ==0:
print("Yes")
else:
print("No")
|
s293673026
|
Accepted
| 17 | 2,940 | 95 |
r,g,b = map(int,input().split())
if (100*r+10*g+b) % 4 == 0:
print("YES")
else:
print("NO")
|
s883799447
|
p03729
|
u337820403
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 123 |
You are given three strings A, B and C. Check whether they form a _word chain_. More formally, determine whether both of the following are true: * The last character in A and the initial character in B are the same. * The last character in B and the initial character in C are the same. If both are true, print `YES`. Otherwise, print `NO`.
|
x = [i for i in input().split()]
torf = [x[0][-1] == x[1][0] and x[1][-1] == x[2][0]]
print("YES" if torf==True else "NO")
|
s634267460
|
Accepted
| 17 | 2,940 | 83 |
a, b, c = input().split()
print("YES" if a[-1] == b[0] and b[-1] == c[0] else "NO")
|
s172745019
|
p03471
|
u887080340
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 426 |
The commonly used bills in Japan are 10000-yen, 5000-yen and 1000-yen bills. Below, the word "bill" refers to only these. According to Aohashi, he received an otoshidama (New Year money gift) envelope from his grandfather that contained N bills for a total of Y yen, but he may be lying. Determine whether such a situation is possible, and if it is, find a possible set of bills contained in the envelope. Assume that his grandfather is rich enough, and the envelope was large enough.
|
N, Y = map(int, input().split())
en_10000 = Y // 10000
if en_10000 > N:
print("-1 -1 -1")
exit()
Y = Y % 10000
N = N - en_10000
en_5000 = Y // 5000
if en_5000 > N:
print("-1 -1 -1")
exit()
Y = Y % 5000
N = N - en_5000
en_1000 = Y // 1000
if en_1000 > N:
print("-1 -1 -1")
exit()
Y = Y % 1000
N = N - en_1000
if (Y > 0) or (N > 0):
print("-1 -1 -1")
exit()
print(en_10000,en_5000,en_1000)
|
s615127911
|
Accepted
| 1,552 | 3,064 | 482 |
N, Y = map(int, input().split())
for i in range((Y // 10000)+1)[::-1]:
if ((Y - 10000 * i) < 0) and ((N - i) < 0):
print("-1 -1 -1")
exit()
Y2 = Y - 10000*i
N2 = N - i
for j in range((Y2 // 5000)+1)[::-1]:
if ((Y2 - 5000 * j) < 0) and ((N2 - j) < 0):
print("-1 -1 -1")
exit()
Y3 = Y2 - 5000 * j
N3 = N2 - j
if N3 * 1000 == Y3:
print(i, j, N3)
exit()
print("-1 -1 -1")
|
s869881671
|
p02420
|
u971748390
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,536 | 160 |
Your task is to shuffle a deck of n cards, each of which is marked by a alphabetical letter. A single shuffle action takes out h cards from the bottom of the deck and moves them to the top of the deck. The deck of cards is represented by a string as follows. abcdeefab The first character and the last character correspond to the card located at the bottom of the deck and the card on the top of the deck respectively. For example, a shuffle with h = 4 to the above deck, moves the first 4 characters "abcd" to the end of the remaining characters "eefab", and generates the following deck: eefababcd You can repeat such shuffle operations. Write a program which reads a deck (a string) and a sequence of h, and prints the final state (a string).
|
while(1):
sen1=input()
if sen1=="-":
break
suff=int(input())
for i in range(suff):
h=int(input())
sen2=sen1[h:]+sen1[:h]
print(sen2)
|
s423708689
|
Accepted
| 30 | 7,656 | 160 |
while(1):
sen1=input()
if sen1=="-":
break
suff=int(input())
for i in range(suff):
h=int(input())
sen1=sen1[h:]+sen1[:h]
print(sen1)
|
s368905861
|
p02613
|
u563217442
| 2,000 | 1,048,576 |
Wrong Answer
| 151 | 15,988 | 242 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
counto = []
for _ in range(int(input())):
s = input()
counto.append(s)
print("AC"+" x "+str(counto.count("AC")))
print("WC"+" x "+str(counto.count("WA")))
print("TLE"+" x "+str(counto.count("TLE")))
print("RE"+" x "+str(counto.count("RE")))
|
s635879156
|
Accepted
| 145 | 9,236 | 246 |
a = 0
w = 0
r = 0
t = 0
for _ in range(int(input())):
s = input()
if s=="AC":
a+=1
elif s=="WA":
w+=1
elif s=="TLE":
t+=1
else:
r+=1
print("AC "+"x "+str(a))
print("WA "+"x "+str(w))
print("TLE "+"x "+str(t))
print("RE "+"x "+str(r))
|
s535389338
|
p03457
|
u835322333
| 2,000 | 262,144 |
Wrong Answer
| 387 | 17,312 | 260 |
AtCoDeer the deer is going on a trip in a two-dimensional plane. In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i. If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1). Note that **he cannot stay at his place**. Determine whether he can carry out his plan.
|
N = int(input())
tmp = [tuple(map(int,input().split())) for i in range(N)]
ct = 0
cx = 0
cy = 0
for t,x,y in tmp:
dst = abs(cx - x) + abs(cy - y)
if dst > t-ct or dst%2 != (t-ct)%2:
print('NO')
exit()
ct,cx,cy = t,x,y
print('YES')
|
s713531798
|
Accepted
| 376 | 17,312 | 256 |
N = int(input())
tmp = [tuple(map(int,input().split())) for i in range(N)]
ct = cx = cy = 0
for t,x,y in tmp:
dst = abs(cx - x) + abs(cy - y)
if dst > t-ct or dst%2 != (t-ct)%2:
print('No')
exit()
ct,cx,cy = t,x,y
print('Yes')
|
s150011780
|
p03160
|
u281573419
| 2,000 | 1,048,576 |
Wrong Answer
| 2,358 | 144,644 | 350 |
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i. There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N: * If the frog is currently on Stone i, jump to Stone i + 1 or Stone i + 2. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on. Find the minimum possible total cost incurred before the frog reaches Stone N.
|
N = int(input())
h = list(map(int, input().split()))
# N = 6
# h = [30, 10, 60, 10, 60, 50]
dp = [0]*(N)
print("ini dp", dp)
for i in range(1, N):
if i == 1:
dp[1] = abs(h[1]-h[0])
print("ini dp 1", dp)
else:
dp[i] = min(dp[i-1]+abs(h[i]-h[i-1]), dp[i-2]+abs(h[i]-h[i-2]))
print("ini dp 2", dp)
print(dp[-1])
|
s031785637
|
Accepted
| 130 | 20,588 | 356 |
N = int(input())
h = list(map(int, input().split()))
# N = 6
# h = [30, 10, 60, 10, 60, 50]
dp = [0]*(N)
for i in range(1, N):
if i == 1:
dp[1] = abs(h[1]-h[0])
else:
dp[i] = min(dp[i-1]+abs(h[i]-h[i-1]), dp[i-2]+abs(h[i]-h[i-2]))
print(dp[-1])
|
s926071984
|
p03438
|
u102960641
| 2,000 | 262,144 |
Wrong Answer
| 27 | 4,596 | 322 |
You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N. Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal. Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions **simultaneously** : * Add 2 to a_i. * Add 1 to b_j.
|
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = []
max_aplus = 0
min_aplus = 0
bplus = 0
for i,j in zip(a,b):
if i > j:
min_aplus += (i-j+1) //2
max_aplus += i-j
else:
bplus += j-i
if bplus >= min_aplus and bplus <= max_aplus:
print("Yes")
else:
print("No")
|
s645636206
|
Accepted
| 45 | 4,664 | 278 |
import math
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = [i-j for i,j in zip(a,b)]
plus = 0
minus = 0
for i in c:
if i >= 0:
plus += i
else:
minus += math.ceil(i / 2)
print("Yes") if abs(minus) >= plus else print("No")
|
s985089807
|
p03386
|
u982594421
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 142 |
Print all the integers that satisfies the following in ascending order: * Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.
|
a, b, k = map(int, input().split())
for i in range(a, min(b, a + k + 1)):
print(i)
for i in range(max(b - k + 1, a + k + 1), b):
print(i)
|
s238826682
|
Accepted
| 17 | 3,060 | 142 |
a, b, k = map(int, input().split())
for i in range(a, min(a + k, b + 1)):
print(i)
for i in range(max(b - k + 1, a + k), b + 1):
print(i)
|
s366283854
|
p03997
|
u525796732
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 278 |
You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively. An example of a trapezoid Find the area of this trapezoid.
|
import sys
stdin = sys.stdin
ns = lambda: stdin.readline().rstrip()
ni = lambda: int(stdin.readline().rstrip())
nm = lambda: map(int, stdin.readline().split())
nl = lambda: list(map(int, stdin.readline().split()))
a=int(input())
b=int(input())
h=int(input())
print((a+b)*h/2)
|
s118573954
|
Accepted
| 17 | 3,060 | 288 |
import sys
stdin = sys.stdin
ns = lambda: stdin.readline().rstrip()
ni = lambda: int(stdin.readline().rstrip())
nm = lambda: map(int, stdin.readline().split())
nl = lambda: list(map(int, stdin.readline().split()))
a=int(input())
b=int(input())
h=int(input())
print(int((a+b)*h/2))
|
s665162499
|
p03546
|
u994521204
| 2,000 | 262,144 |
Wrong Answer
| 36 | 3,444 | 491 |
Joisino the magical girl has decided to turn every single digit that exists on this world into 1. Rewriting a digit i with j (0≤i,j≤9) costs c_{i,j} MP (Magic Points). She is now standing before a wall. The wall is divided into HW squares in H rows and W columns, and at least one square contains a digit between 0 and 9 (inclusive). You are given A_{i,j} that describes the square at the i-th row from the top and j-th column from the left, as follows: * If A_{i,j}≠-1, the square contains a digit A_{i,j}. * If A_{i,j}=-1, the square does not contain a digit. Find the minimum total amount of MP required to turn every digit on this wall into 1 in the end.
|
h, w = map(int, input().split())
C = [list(map(int, input().split())) for i in range(10)]
A = [list(map(int, input().split())) for i in range(h)]
# warshal-floyd
for k in range(10):
for i in range(10):
for j in range(10):
C[i][j] = min(C[i][j], C[i][k] + C[k][j])
ans = 0
for i in range(h):
for j in range(w):
if ans == -1:
ans += 0
else:
ans += C[A[i][j]][1]
print(ans)
|
s677290102
|
Accepted
| 37 | 3,444 | 495 |
h, w = map(int, input().split())
C = [list(map(int, input().split())) for i in range(10)]
A = [list(map(int, input().split())) for i in range(h)]
# warshal-floyd
for k in range(10):
for i in range(10):
for j in range(10):
C[i][j] = min(C[i][j], C[i][k] + C[k][j])
ans = 0
for i in range(h):
for j in range(w):
if A[i][j] == -1:
ans += 0
else:
ans += C[A[i][j]][1]
print(ans)
|
s296680362
|
p03339
|
u384679440
| 2,000 | 1,048,576 |
Wrong Answer
| 218 | 3,672 | 198 |
There are N people standing in a row from west to east. Each person is facing east or west. The directions of the people is given as a string S of length N. The i-th person from the west is facing east if S_i = `E`, and west if S_i = `W`. You will appoint one of the N people as the leader, then command the rest of them to face in the direction of the leader. Here, we do not care which direction the leader is facing. The people in the row hate to change their directions, so you would like to select the leader so that the number of people who have to change their directions is minimized. Find the minimum number of people who have to change their directions.
|
n = int(input())
s = input()
l = 0
r = s[1:].count('E')
result = l + r
for i in range(1, n):
if s[i] == "W":
l += 1
if s[n - i] == "E":
r -= 1
result = max(result, l + r)
print(result)
|
s929253059
|
Accepted
| 215 | 3,700 | 179 |
N = int(input())
S = input()
l = 0
r = S[1:].count('E')
ans = l + r
for i in range(1, N):
if S[i - 1] == 'E':
r -= 1
if S[i] == 'W':
l += 1
ans = min(ans, l + r)
print(ans)
|
s308680767
|
p02413
|
u567380442
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,720 | 256 |
Your task is to perform a simple table calculation. Write a program which reads the number of rows r, columns c and a table of r × c elements, and prints a new table, which includes the total sum for each row and column.
|
r, c = map(int, input().split())
arr = []
for i in range(r):
arr.append(list(map(int, input().split())))
for line in arr:
line.append(sum(line))
arr.append([sum([line[i] for line in arr]) for i in range(c)])
for line in arr:
print(*line)
|
s168607120
|
Accepted
| 50 | 6,844 | 261 |
r, c = map(int, input().split())
arr = []
for i in range(r):
arr.append(list(map(int, input().split())))
for line in arr:
line.append(sum(line))
arr.append([sum([line[i] for line in arr]) for i in range(c + 1)])
for line in arr:
print(*line)
|
s479949391
|
p02255
|
u427088273
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,608 | 236 |
Write a program of the Insertion Sort algorithm which sorts a sequence A in ascending order. The algorithm should be based on the following pseudocode: for i = 1 to A.length-1 key = A[i] /* insert A[i] into the sorted sequence A[0,...,j-1] */ j = i - 1 while j >= 0 and A[j] > key A[j+1] = A[j] j-- A[j+1] = key Note that, indices for array elements are based on 0-origin. To illustrate the algorithms, your program should trace intermediate result for each step.
|
num = int(input())
sort_list = list(map(int,input().split()))
v = 0
for i in range(1,num):
v = sort_list[i]
j = i - 1
while j >= 0 and sort_list[j] > v:
sort_list[j+1] = sort_list[j]
j -= 1
sort_list[j+1] = v
print(*sort_list)
|
s812165064
|
Accepted
| 20 | 7,988 | 254 |
num = int(input())
sort_list = list(map(int,input().split()))
print(*sort_list)
v = 0
for i in range(1,num):
v = sort_list[i]
j = i - 1
while j >= 0 and sort_list[j] > v:
sort_list[j+1] = sort_list[j]
j -= 1
sort_list[j+1] = v
print(*sort_list)
|
s290962178
|
p03110
|
u766566560
| 2,000 | 1,048,576 |
Wrong Answer
| 17 | 2,940 | 161 |
Takahashi received _otoshidama_ (New Year's money gifts) from N of his relatives. You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either `JPY` or `BTC`, and x_i and u_i represent the content of the otoshidama from the i-th relative. For example, if x_1 = `10000` and u_1 = `JPY`, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = `0.10000000` and u_2 = `BTC`, the otoshidama from the second relative is 0.1 bitcoins. If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?
|
ans = 0
N = int(input())
for _ in range(N):
x, y = input().split()
if y == 'JPY':
ans += int(x)
else:
ans += round(float(x) * 380000.0)
print(ans)
|
s932690326
|
Accepted
| 17 | 2,940 | 154 |
ans = 0
N = int(input())
for _ in range(N):
x, y = input().split()
if y == 'JPY':
ans += int(x)
else:
ans += float(x) * 380000.0
print(ans)
|
s149045977
|
p04012
|
u679154596
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,060 | 93 |
Let w be a string consisting of lowercase letters. We will call w _beautiful_ if the following condition is satisfied: * Each lowercase letter of the English alphabet occurs even number of times in w. You are given the string w. Determine if w is beautiful.
|
w = input()
w_set = set(w)
if len(w) / 2 == len(set(w)):
print("Yes")
else:
print("No")
|
s773731552
|
Accepted
| 17 | 2,940 | 197 |
from sys import exit
w = list(input())
w_set = list(set(w))
for i in range(len(w_set)):
if w.count(w_set[i]) % 2 != 0:
print("No")
exit()
else:
if i == len(w_set)-1: print("Yes")
|
s240956880
|
p02613
|
u131411061
| 2,000 | 1,048,576 |
Wrong Answer
| 151 | 16,620 | 236 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
from collections import Counter
N = int(input())
d = {'AC':0,
'WA':0,
'TLE':0,
'RE:':0}
for key,value in Counter([input() for _ in range(N)]).items():
d[key]=value
for key,value in d.items():
print(key,'x',value)
|
s502801497
|
Accepted
| 145 | 16,616 | 235 |
from collections import Counter
N = int(input())
d = {'AC':0,
'WA':0,
'TLE':0,
'RE':0}
for key,value in Counter([input() for _ in range(N)]).items():
d[key]=value
for key,value in d.items():
print(key,'x',value)
|
s223474220
|
p04029
|
u272457181
| 2,000 | 262,144 |
Wrong Answer
| 24 | 9,104 | 34 |
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?
|
N = int(input())
print (N*(N+1)/2)
|
s528669032
|
Accepted
| 26 | 8,972 | 40 |
N = int(input())
print(int((N*(N+1))/2))
|
s412018768
|
p02398
|
u179070318
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 220 |
Write a program which reads three integers a, b and c, and prints the number of divisors of c between a and b.
|
a,b,c = [int(x) for x in input().split( )]
div = set()
for x in range(1,c+1):
if c % x == 0:
div.add(x)
r = set()
for x in range(a,b+1):
r.add(x)
answer_set = div&r
print(' '.join(map(str,answer_set)))
|
s460779053
|
Accepted
| 20 | 5,604 | 137 |
a,b,c = [int(x) for x in input().split( )]
count = 0
for number in range(a,b+1):
if c % number == 0:
count += 1
print(count)
|
s599698514
|
p03455
|
u572122511
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 153 |
AtCoDeer the deer found two positive integers, a and b. Determine whether the product of a and b is even or odd.
|
a = input()
a = a.split()
a = list(map(lambda x: int(x),a))
sum = 1
for i in a:
sum *= i
if sum % 2 == 0:
print("odd")
else:
print("Even")
|
s869058502
|
Accepted
| 17 | 2,940 | 153 |
a = input()
a = a.split()
a = list(map(lambda x: int(x),a))
sum = 1
for i in a:
sum *= i
if sum % 2 == 0:
print("Even")
else:
print("Odd")
|
s415071011
|
p02972
|
u259861571
| 2,000 | 1,048,576 |
Wrong Answer
| 50 | 7,148 | 208 |
There are N empty boxes arranged in a row from left to right. The integer i is written on the i-th box from the left (1 \leq i \leq N). For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it. We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied: * For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2. Does there exist a good set of choices? If the answer is yes, find one good set of choices.
|
# AtCoder
N = int(input())
a = list(map(int, input().split()))
on = -1
if all(a) != True:
print(0)
exit()
sm = sum(a)
if sm % 2 == a[0]:
on = 1
print(1)
print(on)
exit()
print(0)
|
s860513679
|
Accepted
| 561 | 13,104 | 288 |
N = int(input())
A = [0] + list(map(int, input().split()))
B = [0] * (N + 1)
for i in reversed(range(1, N + 1)):
bl = 0
for j in range(i, N + 1, i):
bl += B[j]
if bl % 2 != A[i]:
B[i] = 1
print(sum(B))
print(*[i for i, b in enumerate(B) if b == 1], sep=' ')
|
s278826691
|
p03408
|
u143051858
| 2,000 | 262,144 |
Wrong Answer
| 18 | 3,064 | 257 |
Takahashi has N blue cards and M red cards. A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i. Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen. Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces `atcoder`, he will not earn money even if there are blue cards with `atcoderr`, `atcode`, `btcoder`, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.) At most how much can he earn on balance? Note that the same string may be written on multiple cards.
|
s_num=int(input())
s=[]
s=[input(x) for x in range(s_num)]
t=[]
t_num=int(input())
t=[input(x) for x in range(t_num)]
max_v=0
for i in range(s_num):
if max_v < (s.count(s[i]) - t.count(s[i])):
max_v = (s.count(s[i]) - t.count(s[i]))
print(max_v)
|
s450105985
|
Accepted
| 17 | 3,064 | 364 |
s_num=int(input())
s=[]
for i in range(s_num):
s.append(input())
t=[]
t_num=int(input())
for i in range(t_num):
t.append(input())
max=0
for i in range(s_num):
if max < int(s.count(s[i]) - t.count(s[i])):
max = int(s.count(s[i]) - t.count(s[i]))
#print('int(s.count(s[i]) - t.count(s[i]))=',int(s.count(s[i]) - t.count(s[i])))
print(max)
|
s105619096
|
p03712
|
u044964932
| 2,000 | 262,144 |
Wrong Answer
| 21 | 4,588 | 326 |
You are given a image with a height of H pixels and a width of W pixels. Each pixel is represented by a lowercase English letter. The pixel at the i-th row from the top and j-th column from the left is a_{ij}. Put a box around this image and output the result. The box should consist of `#` and have a thickness of 1.
|
def main():
b, w = map(int, input().split())
image = [list(input()) for _ in range(b)]
print(image)
image.insert(0, ["#"]*w)
image.append(["#"]*w)
for im in image:
im.insert(0, "#")
im.append("#")
for im in image:
print(*im, sep="")
if __name__ == "__main__":
main()
|
s436292657
|
Accepted
| 21 | 4,596 | 309 |
def main():
b, w = map(int, input().split())
image = [list(input()) for _ in range(b)]
image.insert(0, ["#"]*w)
image.append(["#"]*w)
for im in image:
im.insert(0, "#")
im.append("#")
for im in image:
print(*im, sep="")
if __name__ == "__main__":
main()
|
s881951561
|
p02742
|
u819465503
| 2,000 | 1,048,576 |
Wrong Answer
| 320 | 21,060 | 251 |
We have a board with H horizontal rows and W vertical columns of squares. There is a bishop at the top-left square on this board. How many squares can this bishop reach by zero or more movements? Here the bishop can only move diagonally. More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds: * r_1 + c_1 = r_2 + c_2 * r_1 - c_1 = r_2 - c_2 For example, in the following figure, the bishop can move to any of the red squares in one move:
|
import numpy as np
h_w_list = [int(i) for i in input().split()]
if h_w_list[0] == 1 or h_w_list[1] ==1:
print(1)
elif h_w_list[0]*h_w_list[1] % 2 == 0:
print(h_w_list[0]*h_w_list[1]/2)
else:
print(np.ceil(h_w_list[0]*h_w_list[1]/2))
|
s083980505
|
Accepted
| 149 | 12,508 | 261 |
import numpy as np
h_w_list = [int(i) for i in input().split()]
if h_w_list[0] == 1 or h_w_list[1] ==1:
print(1)
elif h_w_list[0]*h_w_list[1] % 2 == 0:
print(int(h_w_list[0]*h_w_list[1]/2))
else:
print(int(np.ceil(h_w_list[0]*h_w_list[1]/2)))
|
s907668745
|
p03399
|
u609814378
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,064 | 207 |
You planned a trip using trains and buses. The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket. Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket. Find the minimum total fare when the optimal choices are made for trains and buses.
|
A = int(input())
B = int(input())
C = int(input())
D = int(input())
A_C = (A+C)
A_D = (A+D)
B_C = (B+C)
B_D = (B+D)
print(max(A_C,A_D,B_C,B_D))
|
s042995898
|
Accepted
| 17 | 3,060 | 207 |
A = int(input())
B = int(input())
C = int(input())
D = int(input())
A_C = (A+C)
A_D = (A+D)
B_C = (B+C)
B_D = (B+D)
print(min(A_C,A_D,B_C,B_D))
|
s704144660
|
p02389
|
u814278309
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,584 | 71 |
Write a program which calculates the area and perimeter of a given rectangle.
|
a, b=map(int, input().split())
print(a, b)
x=a*b
y=2*a+2*b
print(x, y)
|
s239478971
|
Accepted
| 20 | 5,592 | 63 |
a,b=map(int,input().split())
x=a*b
y=2*(a+b)
print(f"{x} {y}")
|
s336687641
|
p03167
|
u439063038
| 2,000 | 1,048,576 |
Wrong Answer
| 657 | 49,360 | 697 |
There is a grid with H horizontal rows and W vertical columns. Let (i, j) denote the square at the i-th row from the top and the j-th column from the left. For each i and j (1 \leq i \leq H, 1 \leq j \leq W), Square (i, j) is described by a character a_{i, j}. If a_{i, j} is `.`, Square (i, j) is an empty square; if a_{i, j} is `#`, Square (i, j) is a wall square. It is guaranteed that Squares (1, 1) and (H, W) are empty squares. Taro will start from Square (1, 1) and reach (H, W) by repeatedly moving right or down to an adjacent empty square. Find the number of Taro's paths from Square (1, 1) to (H, W). As the answer can be extremely large, find the count modulo 10^9 + 7.
|
MOD = 10**9+7
H, W = map(int, input().split())
a = [input() for _ in range(H)]
dp = [[0] * W for _ in range(H)]
dp[0][0] = 1
for i in range(1, H):
if a[i][0] != '#':
dp[i][0] = dp[i-1][0]
else:
dp[i][0] = 0
for i in range(1, W):
if a[0][i] != '#':
dp[0][i] = dp[0][i-1]
else:
dp[0][i] = 0
for i in range(1, H):
for j in range(1, W):
if a[i][j] == '#':
continue
if a[i-1][j] != '#' and a[i][j-1] != '#':
dp[i][j] = (dp[i-1][j] + dp[i][j-1]) % MOD
elif a[i-1][j] != '#':
dp[i][j] = dp[i-1][j]
elif a[i][j-1] != '#':
dp[i][j] = dp[i-1][j]
print(dp[H-1][W-1])
|
s175977862
|
Accepted
| 577 | 51,216 | 382 |
H, W = map(int, input().split())
a = [input() for _ in range(H)]
MOD = 10**9 + 7
dp = [[0]*1100 for _ in range(1100)]
dp[1][1] = 1
for h in range(1, H+1):
for w in range(1, W+1):
if h == 1 and w == 1:
continue
if a[h-1][w-1] == '#':
dp[h][w] = 0
else:
dp[h][w] += (dp[h-1][w] + dp[h][w-1]) % MOD
print(dp[H][W] % MOD)
|
s889396965
|
p02612
|
u475189661
| 2,000 | 1,048,576 |
Wrong Answer
| 28 | 9,024 | 61 |
We will buy a product for N yen (the currency of Japan) at a shop. If we use only 1000-yen bills to pay the price, how much change will we receive? Assume we use the minimum number of bills required.
|
n = int(input())
b = 0
while b > n:
b += 1000
print(b-n)
|
s422484134
|
Accepted
| 26 | 9,056 | 58 |
n = int(input())
b = 0
while b < n:
b += 1000
print(b-n)
|
s279888407
|
p02397
|
u326248180
| 1,000 | 131,072 |
Wrong Answer
| 20 | 7,556 | 104 |
Write a program which reads two integers x and y, and prints them in ascending order.
|
x, y = map(int, input().split())
if x < y:
print("%d %d" % (x, y))
else:
print("%d %d" % (y, x))
|
s546250943
|
Accepted
| 40 | 7,660 | 242 |
import sys
while(True):
x, y = map(lambda x: (int, int)[x.isdigit()](x) ,sys.stdin.readline().split(None, 1))
if x == 0 and y == 0:
break
if x < y:
print("%d %d" % (x, y))
else:
print("%d %d" % (y, x))
|
s998780063
|
p03545
|
u103902792
| 2,000 | 262,144 |
Wrong Answer
| 17 | 3,060 | 291 |
Sitting in a station waiting room, Joisino is gazing at her train ticket. The ticket is numbered with four digits A, B, C and D in this order, each between 0 and 9 (inclusive). In the formula A op1 B op2 C op3 D = 7, replace each of the symbols op1, op2 and op3 with `+` or `-` so that the formula holds. The given input guarantees that there is a solution. If there are multiple solutions, any of them will be accepted.
|
s = input()
def func(ac,s,op):
next = s[0]
if len(s) ==1:
if ac + next == 7:
print(str+'+'+s[0]+'=7')
exit(0)
if ac - next == 7:
print(str+'-'+s[0]+'=7')
exit(0)
else:
func(ac+next,s[1:],op+'+'+str(next))
func(ac-next,s[1:],op+'-'+str(next))
|
s598593239
|
Accepted
| 17 | 3,060 | 323 |
s = input()
def func(ac,s,op):
next = int(s[0])
if len(s) ==1:
if ac + next == 7:
print(op+'+'+s[0]+'=7')
exit(0)
if ac - next == 7:
print(op+'-'+s[0]+'=7')
exit(0)
else:
func(ac+next,s[1:],op+'+'+str(next))
func(ac-next,s[1:],op+'-'+str(next))
func(int(s[0]),s[1:],s[0])
|
s006608123
|
p03477
|
u708019102
| 2,000 | 262,144 |
Wrong Answer
| 17 | 2,940 | 140 |
A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R. Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan. Print `Left` if the balance scale tips to the left; print `Balanced` if it balances; print `Right` if it tips to the right.
|
a,b,c,d = [int(x) for x in input().split()]
r = a+b
l = c+d
if r > l:
print("Right")
elif l > r:
print("Left")
else:
print("Balanced")
|
s184705879
|
Accepted
| 18 | 3,060 | 140 |
a,b,c,d = [int(x) for x in input().split()]
l = a+b
r = c+d
if r > l:
print("Right")
elif l > r:
print("Left")
else:
print("Balanced")
|
s510594314
|
p03944
|
u292810930
| 2,000 | 262,144 |
Wrong Answer
| 149 | 12,508 | 538 |
There is a rectangle in the xy-plane, with its lower left corner at (0, 0) and its upper right corner at (W, H). Each of its sides is parallel to the x-axis or y-axis. Initially, the whole region within the rectangle is painted white. Snuke plotted N points into the rectangle. The coordinate of the i-th (1 ≦ i ≦ N) point was (x_i, y_i). Then, he created an integer sequence a of length N, and for each 1 ≦ i ≦ N, he painted some region within the rectangle black, as follows: * If a_i = 1, he painted the region satisfying x < x_i within the rectangle. * If a_i = 2, he painted the region satisfying x > x_i within the rectangle. * If a_i = 3, he painted the region satisfying y < y_i within the rectangle. * If a_i = 4, he painted the region satisfying y > y_i within the rectangle. Find the area of the white region within the rectangle after he finished painting.
|
import numpy as np
W, H, N = map(int, input().split())
xlist = np.zeros(N,dtype=int)
ylist = np.zeros(N,dtype=int)
alist = np.zeros(N,dtype=int)
for i in range(N):
xlist[i], ylist[i], alist[i] = map(int, input().split())
rectangular = np.zeros((H, W))
for i in range(N):
if alist[i] == 1:
rectangular[:,:xlist[i]] = 1
if alist[i] == 2:
rectangular[:,xlist[i]:] = 1
if alist[i] == 3:
rectangular[:ylist[i],:] = 1
if alist[i] == 4:
rectangular[ylist[i]:,:] = 1
sum(sum(rectangular < 1))
|
s805167505
|
Accepted
| 150 | 12,508 | 545 |
import numpy as np
W, H, N = map(int, input().split())
xlist = np.zeros(N,dtype=int)
ylist = np.zeros(N,dtype=int)
alist = np.zeros(N,dtype=int)
for i in range(N):
xlist[i], ylist[i], alist[i] = map(int, input().split())
rectangular = np.zeros((H, W))
for i in range(N):
if alist[i] == 1:
rectangular[:,:xlist[i]] = 1
if alist[i] == 2:
rectangular[:,xlist[i]:] = 1
if alist[i] == 3:
rectangular[:ylist[i],:] = 1
if alist[i] == 4:
rectangular[ylist[i]:,:] = 1
print(sum(sum(rectangular < 1)))
|
s176609741
|
p02409
|
u444997287
| 1,000 | 131,072 |
Wrong Answer
| 30 | 7,712 | 413 |
You manage 4 buildings, each of which has 3 floors, each of which consists of 10 rooms. Write a program which reads a sequence of tenant/leaver notices, and reports the number of tenants for each room. For each notice, you are given four integers b, f, r and v which represent that v persons entered to room r of fth floor at building b. If v is negative, it means that −v persons left. Assume that initially no person lives in the building.
|
N = int(input())
m=[]
line = ''
m = [[[0 for k in range(10)] for j in range(3)] for i in range(10)]
for i in range(N):
n = input().split()
b = int(n[0])
f = int(n[1])
r = int(n[2])
v = int(n[3])
m[b-1][f-1][r-1] += v
for i in range(4):
for j in range(3):
for k in range(10):
line += ' {0}'.format(m[i][j][k])
print(line)
line = ''
print('#'*20)
|
s067092519
|
Accepted
| 20 | 7,716 | 431 |
N = int(input())
m=[]
line = ''
m = [[[0 for k in range(10)] for j in range(3)] for i in range(10)]
for i in range(N):
n = input().split()
b = int(n[0])
f = int(n[1])
r = int(n[2])
v = int(n[3])
m[b-1][f-1][r-1] += v
for i in range(4):
for j in range(3):
for k in range(10):
line += ' {0}'.format(m[i][j][k])
print(line)
line = ''
if i < 3:
print('#'*20)
|
s826968930
|
p02261
|
u056253438
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,604 | 1,062 |
Let's arrange a deck of cards. There are totally 36 cards of 4 suits(S, H, C, D) and 9 values (1, 2, ... 9). For example, 'eight of heart' is represented by H8 and 'one of diamonds' is represented by D1. Your task is to write a program which sorts a given set of cards in ascending order by their values using the Bubble Sort algorithms and the Selection Sort algorithm respectively. These algorithms should be based on the following pseudocode: BubbleSort(C) 1 for i = 0 to C.length-1 2 for j = C.length-1 downto i+1 3 if C[j].value < C[j-1].value 4 swap C[j] and C[j-1] SelectionSort(C) 1 for i = 0 to C.length-1 2 mini = i 3 for j = i to C.length-1 4 if C[j].value < C[mini].value 5 mini = j 6 swap C[i] and C[mini] Note that, indices for array elements are based on 0-origin. For each algorithm, report the stability of the output for the given input (instance). Here, 'stability of the output' means that: cards with the same value appear in the output in the same order as they do in the input (instance).
|
_, cards = input(), input().split()
def bubble_sort(array):
A = array[:]
for i in range(len(A)):
for j in range(1, len(A) - i):
if int(A[j - 1][1]) > int(A[j][1]):
A[j], A[j - 1] = A[j - 1], A[j]
return A
def selection_sort(array):
A = array[:]
for i in range(len(A)):
minn = i
for j in range(i, len(A)):
if int(A[j][1]) < int(A[minn][1]): minn = j
A[i], A[minn] = A[minn], A[i]
return A
def is_stable(org, sorted):
def get_same_numbers(array):
result = {}
for i in range(1, 10):
result[i] = []
for j in range(len(array)):
if int(array[j][1]) == i:
result[i].append(array[j])
return [cards for _, cards in result.items() if len(cards) >= 2]
return "Stable" if get_same_numbers(org) == get_same_numbers(sorted) else "Not Stable"
a = bubble_sort(cards)
print(" ".join(a))
print(is_stable(cards, a))
b = selection_sort(cards)
print(" ".join(b))
print(is_stable(cards, b))
|
s733008742
|
Accepted
| 20 | 5,612 | 1,062 |
_, cards = input(), input().split()
def bubble_sort(array):
A = array[:]
for i in range(len(A)):
for j in range(1, len(A) - i):
if int(A[j - 1][1]) > int(A[j][1]):
A[j], A[j - 1] = A[j - 1], A[j]
return A
def selection_sort(array):
A = array[:]
for i in range(len(A)):
minn = i
for j in range(i, len(A)):
if int(A[j][1]) < int(A[minn][1]): minn = j
A[i], A[minn] = A[minn], A[i]
return A
def is_stable(org, sorted):
def get_same_numbers(array):
result = {}
for i in range(1, 10):
result[i] = []
for j in range(len(array)):
if int(array[j][1]) == i:
result[i].append(array[j])
return [cards for _, cards in result.items() if len(cards) >= 2]
return "Stable" if get_same_numbers(org) == get_same_numbers(sorted) else "Not stable"
a = bubble_sort(cards)
print(" ".join(a))
print(is_stable(cards, a))
b = selection_sort(cards)
print(" ".join(b))
print(is_stable(cards, b))
|
s340073746
|
p02613
|
u339194100
| 2,000 | 1,048,576 |
Wrong Answer
| 163 | 9,224 | 283 |
Takahashi is participating in a programming contest called AXC002, and he has just submitted his code to Problem A. The problem has N test cases. For each test case i (1\leq i \leq N), you are given a string S_i representing the verdict for that test case. Find the numbers of test cases for which the verdict is `AC`, `WA`, `TLE`, and `RE`, respectively. See the Output section for the output format.
|
n=int(input())
a=0
w=0
t=0
r=0
for i in range(n):
s=str(input())
if(s=='AC'):
a+=1
elif(s=="WA"):
w+=1
elif(s=="TLE"):
t+=1
else:
r+=1
print('AC ', '×　',a)
print('WA ', '×　',w)
print('TLE ', '×　',t)
print('RE ', '×　',r)
|
s675297311
|
Accepted
| 162 | 9,220 | 263 |
n=int(input())
a=0
w=0
t=0
r=0
for i in range(n):
s=str(input())
if(s=='AC'):
a+=1
elif(s=="WA"):
w+=1
elif(s=="TLE"):
t+=1
else:
r+=1
print('AC', 'x',a)
print('WA', 'x',w)
print('TLE', 'x',t)
print('RE', 'x',r)
|
s964660572
|
p02288
|
u024715419
| 2,000 | 131,072 |
Wrong Answer
| 20 | 7,684 | 94 |
A binary heap which satisfies max-heap property is called max-heap. In a max- heap, for every node $i$ other than the root, $A Write a program which reads an array and constructs a max-heap from the array based on the following pseudo code. $maxHeapify(A, i)$ move the value of $A[i]$ down to leaves to make a sub-tree of node $i$ a max-heap. Here, $H$ is the size of the heap. 1 maxHeapify(A, i) 2 l = left(i) 3 r = right(i) 4 // select the node which has the maximum value 5 if l ≤ H and A[l] > A[i] 6 largest = l 7 else 8 largest = i 9 if r ≤ H and A[r] > A[largest] 10 largest = r 11 12 if largest ≠i　// value of children is larger than that of i 13 swap A[i] and A[largest] 14 maxHeapify(A, largest) // call recursively The following procedure buildMaxHeap(A) makes $A$ a max-heap by performing maxHeapify in a bottom-up manner. 1 buildMaxHeap(A) 2 for i = H/2 downto 1 3 maxHeapify(A, i)
|
h = int(input())
key = list(map(int,input().split()))
key.sort()
key.reverse()
print("",*key)
|
s143925166
|
Accepted
| 910 | 65,996 | 451 |
def maxHeapify(a, i):
l = 2*i
r = 2*i + 1
if l <= h and a[l] > a[i]:
largest = l
else:
largest = i
if r <= h and a[r] > a[largest]:
largest = r
if largest != i:
a[i], a[largest] = a[largest], a[i]
maxHeapify(a, largest)
h = int(input())
n = int(0.5*h)
key = list(map(int,input().split()))
key.insert(0,0)
for i in range(n):
j = n - i
maxHeapify(key, j)
print("",*key[1:])
|
s865411034
|
p03761
|
u202570162
| 2,000 | 262,144 |
Wrong Answer
| 19 | 3,064 | 294 |
Snuke loves "paper cutting": he cuts out characters from a newspaper headline and rearranges them to form another string. He will receive a headline which contains one of the strings S_1,...,S_n tomorrow. He is excited and already thinking of what string he will create. Since he does not know the string on the headline yet, he is interested in strings that can be created regardless of which string the headline contains. Find the longest string that can be created regardless of which string among S_1,...,S_n the headline contains. If there are multiple such strings, find the lexicographically smallest one among them.
|
alp=list('qawsedrftgyhujikoplmnbvcxz')
alp.sort()
cnt=[]
for i in range(int(input())):
tmp=[0 for _ in range(26)]
for s in list(input()):
tmp[alp.index(s)]+=1
cnt.append(tmp)
m=57
ans=''
for i in range(26):
for j in range(len(cnt)):
m=min(m,cnt[j][i])
ans+=alp[i]*m
print(ans)
|
s598291516
|
Accepted
| 19 | 3,064 | 325 |
alp=list('qawsedrftgyhujikoplmnbvcxz')
alp.sort()
# print(alp)
cnt=[]
for i in range(int(input())):
tmp=[0 for _ in range(26)]
for s in list(input()):
tmp[alp.index(s)]+=1
cnt.append(tmp)
# print(tmp)
ans=''
for i in range(26):
m=57
for j in range(len(cnt)):
m=min(m,cnt[j][i])
ans+=alp[i]*m
print(ans)
|
s133460154
|
p02412
|
u971748390
| 1,000 | 131,072 |
Wrong Answer
| 30 | 6,724 | 216 |
Write a program which identifies the number of combinations of three integers which satisfy the following conditions: * You should select three distinct integers from 1 to n. * A total sum of the three integers is x. For example, there are two combinations for n = 5 and x = 9. * 1 + 3 + 5 = 9 * 2 + 3 + 4 = 9
|
while True:
n,x=map(int,input().split())
if n==x==0: break
count=0
for i in range(n-2):
for k in range(i+1,n-1):
for j in range(k+1,n):
if i+k+j==x :
count = count +1
print('count')
|
s630592933
|
Accepted
| 640 | 6,724 | 234 |
while True:
n,x=map(int,input().split())
if n==x==0 : break
count = 0
for i in range(1,n-1) :
for j in range(i+1,n):
for k in range(j+1,n+1):
if i+j+k==x :
count = count +1
print(count)
|
s771856907
|
p03401
|
u953110527
| 2,000 | 262,144 |
Wrong Answer
| 227 | 14,800 | 383 |
There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis. You planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0. However, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned. For each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.
|
n = int(input())
c = list(map(int,input().split()))
a = [0]
a += c
a.append(0)
b = [0 for i in range(n+1)]
b[0] = abs(a[1])
count = abs(a[1])
for i in range(1,n):
b[i] = abs(a[i+1]-a[i])
count += b[i]
count += abs(a[n])
b[n] = abs(a[n])
print(a)
print(b)
for i in range(n):
print(count - b[i] - b[i+1] + abs(a[i+2] - a[i]))
# 0 3 5 -1 0
# 0 5 -1 0
# 0 3 -1 0
# 0 3 5 0
|
s523681335
|
Accepted
| 201 | 14,048 | 365 |
n = int(input())
c = list(map(int,input().split()))
a = [0]
a += c
a.append(0)
b = [0 for i in range(n+1)]
b[0] = abs(a[1])
count = abs(a[1])
for i in range(1,n):
b[i] = abs(a[i+1]-a[i])
count += b[i]
count += abs(a[n])
b[n] = abs(a[n])
for i in range(n):
print(count - b[i] - b[i+1] + abs(a[i+2] - a[i]))
# 0 3 5 -1 0
# 0 5 -1 0
# 0 3 -1 0
# 0 3 5 0
|
s310224806
|
p02262
|
u387437217
| 6,000 | 131,072 |
Wrong Answer
| 20 | 7,820 | 660 |
Shell Sort is a generalization of [Insertion Sort](http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ALDS1_1_A) to arrange a list of $n$ elements $A$. 1 insertionSort(A, n, g) 2 for i = g to n-1 3 v = A[i] 4 j = i - g 5 while j >= 0 && A[j] > v 6 A[j+g] = A[j] 7 j = j - g 8 cnt++ 9 A[j+g] = v 10 11 shellSort(A, n) 12 cnt = 0 13 m = ? 14 G[] = {?, ?,..., ?} 15 for i = 0 to m-1 16 insertionSort(A, n, G[i]) A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every $g$-th elements. Beginning with large values of $g$, it repeats the insertion sort with smaller $g$. Your task is to complete the above program by filling ?. Write a program which reads an integer $n$ and a sequence $A$, and prints $m$, $G_i (i = 0, 1, ..., m − 1)$ in the pseudo code and the sequence $A$ in ascending order. The output of your program must meet the following requirements: * $1 \leq m \leq 100$ * $0 \leq G_i \leq n$ * cnt does not exceed $\lceil n^{1.5}\rceil$
|
# coding: utf-8
# Here your code !
import math
n=int(input())
A=[int(input()) for i in range(n)]
def insertion_sort(ar,num,g):
count=0
for i in range(g,num):
v=ar[i]
j=i-g
while j>=0 and ar[j]>v:
ar[j+g]=ar[j]
j=j-g
count+=1
ar[j+g]=v
return count
def shell_sort(AR,num):
count=0
m = math.floor(math.log(2 * n + 1) / math.log(3))
print(m)
G = [0 for i in range(m)]
G[0] = 1
for i in range(1, m):
G[i] = 3 * G[i - 1] + 1
print(*G)
for i in range(m):
insertion_sort(AR,n,G[i])
shell_sort(A,n)
for i in A:
print(i)
|
s689029847
|
Accepted
| 20,880 | 47,524 | 670 |
import math
n=int(input())
A=[int(input()) for i in range(n)]
def insertion_sort(ar,num,g):
count=0
for i in range(g,num):
v=ar[i]
j=i-g
while j>=0 and ar[j]>v:
ar[j+g]=ar[j]
j=j-g
count+=1
ar[j+g]=v
return count
def shell_sort(AR,num):
count=0
m = math.floor(math.log(2 * n + 1) / math.log(3))
print(m)
G = [0 for i in range(m)]
G[0] = 1
for i in range(1, m):
G[i] = 3 * G[i - 1] + 1
G.reverse()
print(*G)
for i in range(m):
count+= insertion_sort(A, n, G[i])
print(count)
shell_sort(A,n)
for i in A:
print(i)
|
s732273078
|
p02392
|
u344890307
| 1,000 | 131,072 |
Wrong Answer
| 20 | 5,600 | 157 |
Write a program which reads three integers a, b and c, and prints "Yes" if a < b < c, otherwise "No".
|
def main():
a,b,c=map(int,input().split())
if a<b<c:
print('yes')
else:
print('No')
if __name__=='__main__':
main()
|
s045589874
|
Accepted
| 20 | 5,600 | 157 |
def main():
a,b,c=map(int,input().split())
if a<b<c:
print('Yes')
else:
print('No')
if __name__=='__main__':
main()
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.