TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
from itertools import (
    accumulate,  
    groupby,  
    permutations,  
    combinations,  
    product,  
    combinations_with_replacement,  
)

import fractions
from functools import reduce

def gcd(*numbers):
    return reduce(fractions.gcd, numbers)

def gcd_list(numbers):
    return reduce(fractions.gcd, numbers)


N = int(input())
ans = 0
for v in product(range(1, N+1), repeat=3):
    ans += gcd_list(v)

print(ans)

from itertools import (
    accumulate,  
    groupby,  
    permutations,  
    combinations,  
    product,  
    combinations_with_replacement,  
)

import fractions
from functools import reduce

def gcd(*numbers):
    return reduce(fractions.gcd, numbers)

def gcd_list(numbers):
    return reduce(fractions.gcd, numbers)


N = int(input())
ans = 0
for i, j, k in combinations_with_replacement(range(1, N+1), 3):
    if i==j==k:
        ans += i
    elif i==j:
        ans += gcd(i, k)*3
    elif j==k:
        ans += gcd(j, i)*3
    elif i==k:
        ans += gcd(i, j)*3
    else:
        ans += gcd_list([i, j, k])*6
print(ans)

from itertools import (
    accumulate,  
    groupby,  
    permutations,  
    combinations,  
    product,  
    combinations_with_replacement,  
)

import fractions, math
from functools import reduce

def gcd_list(numbers):
    return reduce(math.gcd, numbers)


N = int(input())
ans = 0
for i, j, k in combinations_with_replacement(range(1, N+1), 3):
    if i==j==k:
        ans += i
    elif i==j:
        ans += math.gcd(i, k)*3
    elif j==k:
        ans += math.gcd(j, i)*3
    elif i==k:
        ans += math.gcd(i, j)*3
    else:
        ans += gcd_list([i, j, k])*6
print(ans)

Verdicts: Time Limit Exceeded

Time Limit Exceeded

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
k=int(input())+1
from math import gcd
print(sum(gcd(gcd(i,j),r) for i in range(1,k)   for j in range(1,k)  for r in range(1,k) )
k=int(input())+1
from math import gcd
print(sum( gcd(gcd(i,j),r) for i in range(1,k)   for j in range(1,k)  for r in range(1,k) ))


k=int(input())+1
from math import gcd
print(sum(gcd(gcd(i,j),r) for i in range(1,k)   for j in range(1,k)  for r in range(1,k) )


k=int(input())+1
from math import gcd
print(sum( gcd(gcd(i,j),r) for i in range(1,k)   for j in range(1,k)  for r in range(1,k) ))


Verdicts: Runtime Error

Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import io


def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)


def __main__():
    K = int(input())

    ans = 0
    for a in range(1,K+1):
        for b in range(1,K+1):
            for c in range(1,K+1):
                ans += gcd(c, gcd(a, b))

    print(ans)

import numpy as np


def gcd(a, b):
    while b > 0 :
        a, b = b, a%b

    return a


K = int(input())
ans = 0
for a in range(1,K+1):
    for b in range(1,K+1):
        g = gcd(a, b)
        for c in range(1,K+1):
            ans += gcd(c, g)

print(ans)

Verdicts: Wrong Answer

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
n=int(input())
import math
ans=0
for i in range(2,n+1):
    for j in range(1,n+1):
        ij=math.gcd(i,j)
        if ij==1:
            ans+=n
        else:
            for k in range(1,n+1):
                ans+=math.gcd(ij,k)
print(ans)

n=int(input())
import math
ans=0
for i in range(1,n+1):
    for j in range(1,n+1):
        ij=math.gcd(i,j)
        if ij==1:
            ans+=n
        else:
            for k in range(1,n+1):
                ans+=math.gcd(ij,k)
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
def main():
    k = int(input())
    gcds = list(sorted([gcd(i, j) for i in range(1, k+1) for j in range(1, k+1)]))
    j = 0
    s = list(set(gcds))
    m = k**2-1
    gcdss = [0]*len(s)
    for i, l in enumerate(s):
        while gcds[j] == l:
            j += 1
            gcdss[i] += 1
        if j == m:
            break
    ans = sum([gcd(i, j)*n for i in range(1, k+1) for j, n in zip(s, gcdss)])
    print(ans)
            
    
def gcd(a, b):
    if a < b:
        a, b = b, a
    if a % b == 0:
        return b
    else:
        return gcd(b, a%b)


if __name__ == "__main__":
    main()

def main():
    k = int(input())
    gcds = []
    for i in range(1, k+1):
        for j in range(1, k+1):
            gcds.append(gcd(i, j))
    ans = {i:0 for i in set(gcds)}
    for i in gcds:
        ans[i] += 1
    out = 0
    for i, v in ans.items():
        for j in range(1, k+1):
            out += gcd(i, j)*v
    print(out)

def gcd(x, y):
    r = x % y
    while r:
        x, y, r = y, r, y%r
    return y

if __name__ == "__main__":
    main()

Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math
from functools import reduce

def gcd(*numbers):
    return reduce(math.gcd, numbers)
  
k=int(input())
r=0
l=[]

for i in range(k):
  for j in range(k):
    for l in range(k):
      if i<=j and j<=k:
        l.append([i+1,j+1,k+1])
for s in l:
  r+=gcd(s[0],s[1],s[2])
print(r)


import math
  
k=int(input())
r=0
ls=[]

for i in range(k):
  for j in range(k):
    a=math.gcd(i+1,j+1)
    for l in range(k):
      r+=math.gcd(a,l+1)
print(r)

Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
a=str(input())
if '7' in a:
    print("Yes")
else:
    print("No")

from math import gcd
k=int(input())
re=0
for i in range(1,k+1):
    for j in range(1,k+1):
        v=gcd(i,j)
        for l in range(1,k+1):
            re+=gcd(v,l)
print(re)


Verdicts: Wrong Answer

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math

k = int(input())
ans = 0
for a in range(1, k+1):
    for b in range(1, k+1):
        ab = math.gcd(a, b)
        if ab == 1:
            ans += k
            continue
    for k in range(1, k+1):
        ans += math.gcd(ab,k)

print(ans)


import math

k = int(input())
ans = 0
for a in range(1, k+1):
    for b in range(1, k+1):
        ab = math.gcd(a, b)
        if ab == 1:
            ans += k
            continue
        for c in range(1, k+1):
            ans += math.gcd(ab,c)

print(ans)

Verdicts: Wrong Answer

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math
from functools import reduce
  
k = int(input())
k+=1

sum = 0
for c in range(1,k):
  for b in range(1,k):
    for a in range(1,k):
      sum+=reduce(math.gcd, [a,b,c])

import math
from functools import reduce
  
k = int(input())
k+=1

sum = 0
for a in range(1,k):
  for b in range(a,k):
    for c in range(b,k):
      t = len(set([a,b,c]))
      if t==1:
        sum+= reduce(math.gcd, [a,b,c])
      elif t==2:
        sum+= 3*reduce(math.gcd, [a,b,c])
      else:
        sum+= 6*reduce(math.gcd, [a,b,c])

print(sum)

Verdicts: Wrong Answer

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import fractions

K = int(input())
result = 0
sum = 0

for a in range(1,K+1):
    for b in range(1,K+1):
        for c in range(1,K+1):
            result = fractions.gcd(a, b)
            result = fractions.gcd(result, c)
            sum = sum + result

print(sum)  

import math
K = int(input())
result = 0
sum = 0

for a in range(1,K+1):
    for b in range(1,K+1):
        result = math.gcd(a, b)
        for c in range(1,K+1):
            sum += math.gcd(result, c)

print(sum)            


Verdicts: Time Limit Exceeded

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
k = int(input())

sum = 0
for a in range(1, k+1):
    for b in range(1, k+1):
        for c in range(1,k+1):
            sum += gcd(gcd(a,b), c)

print(sum)

def gcd(a, b):
    while b:
        a, b = b, a % b
    return a

def resolve():
    k = int(input())

    tmp = []
    for a in range(1, k+1):
        for b in range(1, k+1):
            tmp.append(gcd(a,b))

    ans = 0
    for c in range(1,k+1):
        for i in tmp:
            ans += gcd(i, c)
    
    print(ans)
    
resolve()

Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math
K = int(input())
sum = 0
for a in range(1,K+1):
    for b in range(1,K+1):
        for c in range(1,K+1):
            ans+=math.gcd(math.gcd(a,b),c)
print(sum(list))


import math
K = int(input())
sum = 0
for a in range(1,K+1):
    for b in range(1,K+1):
        x = math.gcd(a,b)
        for c in range(1,K+1):
            sum+=math.gcd(x,c)
print(sum)

Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math
from functools import reduce
import itertools
import numpy as np


def gcd_list(numbers):
    return reduce(math.gcd, numbers)


K = int(input())

l = [i for i in range(1, K + 1)]

temp_1 = np.array([])
for j in itertools.combinations(l, 3):
    temp_1=np.append(temp_1,gcd_list(list(j)))

a=np.sum(temp_1)*6

temp_2 = np.array([])

for j in itertools.combinations(l, 2):
    temp_2=np.append(temp_2,gcd_list(list(j[0],j[0],j[1])))
    temp_2=np.append(temp_2,gcd_list(list(j[0],j[1],j[1])))
    
b=np.sum(temp_2)*3

temp_3 = np.array([])
for i in l:
    temp_3 = np.append(temp_3, gcd_list([i, i, i]))

print(int(np.sum(temp_3)+a+b))


import math
from functools import reduce
import itertools
import numpy as np

K = int(input())



ans = 0
for i in range(1, K + 1):
    for j in range(i, K + 1):
        for k in range(j, K + 1):
            if i == j and j == k:
                dup_num = 1
            elif i == j or j == k:
                dup_num = 3
            else:
                dup_num = 6
            ans += math.gcd(math.gcd(i, j), k) * dup_num

print(ans)


Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math
k=int(input())
list=[]
for a in range(1,k+1):
  for b in range(1,k+1):
    for c in range(1,k+1):
      list.append(math.gcd((a,b),c))
print(sum(list))

import math
k=int(input())
list=[]
for a in range(1,k+1):
  for b in range(1,k+1):
    for c in range(1,k+1):
      list.append(math.gcd(gcd(a,b),c))
print(sum(list))

import math
k=int(input())
list=[]
for a in range(1,k+1):
  for b in range(1,k+1):
    t=math.gcd(a,b)
    for c in range(1,k+1):
      list.append(math.gcd(t,c))
print(sum(list))

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
from math import math

k = int(input())
count = 0
for i in range(1,k+1):
    for j in range(1,k+1):
        for l in range(1,k+1):
            count += gcd(gcd(j,l),i)
print(count)


from math import gcd
K=int(input())
ans=0 
 
for i in range(1,K+1):
  for j in range(1,K+1):
    for l in range(1,K+1):
      ans+= gcd(gcd(l,j),i)
 
print(ans)


Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
n=int(input())
a=0
for i in range(n+1):
    for j in range(n+1):
        for k in range(n+1):
            a+=g[g[i][j]][k]
print(a)

from math import *
g=[[0]*202 for _ in range(202)]
for i in range(1,201):
    for j in range(1,201):
        g[i][j]=gcd(i,j)
n=int(input())
a=0
for i in range(n+1):
    for j in range(n+1):
        for k in range(n+1):
            a+=g[g[i][j]][k]
print(a)

Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
n = int(input())
result = [0]*n**3
for a in range(n):
    for b in range(n):
        for c in range(n):
            if a > b:
                result[n*n*a +n*b + c]= result[n*n*b +n*a + c]
            elif b > c:
                result[n*n*a +n*b + c]= result[n*n*a +n*c + b]
            else:
                result[n*n*a +n*b + c] = gcd(a+1, gcd(b+1,c+1))
print(sum(result))

n = int(input())
answer = 0

def gcd(i, j):
    while j:
        i, j = j, i % j
    return i

for a in range(1,n+1):
    for b in range(1,n+1):
        d = gcd(a,b)
        for c in range(1, n+1):
            answer += gcd(c, d)
print(answer)

Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
from fractions import gcd
k=int(input())
sum=0
for l in range(1,k+1):
    for m in range(1,k+1):
        for n in range(1,k+1):
            sum += gcd(gcd(l,m),n)

print(sum)

from fractions import gcd
k=int(input())
r1=[]
r2=[]
for l in range(1,k+1):
    for m in range(1,k+1):
        r1.append(gcd(l,m))

for i in range(len(r1)):
    for j in range(1,k+1):
        r2.append(gcd(r1[i],j))

print(sum(r2))

from math import gcd
k=int(input())

sum=0

for i in range(1,k+1):
    for j in range(1,k+1):
        for l in range(1,k+1):
            sum += gcd(gcd(i,j),l)

print(sum)

Verdicts: Time Limit Exceeded

Time Limit Exceeded

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math

K=2
count = 0
for a in range(1,K+1):
    for b in range(1,a+1):
        for c in range(1,b+1):
            print(a,b,c)
            _z=math.gcd(a,b)
            if a==b and b==c:
                k=1
            else:
                k=3
            count += k * math.gcd(_z,c)
print(count)


import math

K=int(input())
count = 0
for a in range(1,K+1):
  for b in range(1,a+1):
    for c in range(1,b+1):
      _z=math.gcd(a,b)
      count += math.gcd(_z,c)
print(count)

import math

K=int(input())
count = 0
for a in range(1,K+1):
    for b in range(1,a+1):
        for c in range(1,b+1):
            _z=math.gcd(a,b)
            if a==b and b==c:
                k=1
            elif a==b or b==c or c==a:
                k=3
            else:
                k=6
            count += k * math.gcd(_z,c)
print(count)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math
K = int(input())
count = 0
for i in range(1,K+1):
    for j in range(i,K+1):
        for k in range(1,K+1):
            if i != j:
                count += math.gcd(math.gcd(i,j),k)
            else:
                count += math.gcd(math.gcd(i,j),k)
print(count)

import math
K = int(input())
count = 0
for i in range(1,K+1):
    for j in range(i,K+1):
        for k in range(1,K+1):
            if i != j:
                count += 2 * math.gcd(math.gcd(i,j),k)
            else:
                count += math.gcd(math.gcd(i,j),k)
print(count)

Verdicts: Wrong Answer

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
from math import gcd
key = int(input())
sin = 0
for i in range(1,k+1):
    for j in range(1,k+1):
        tmp = gcd(i,j)
        for m in range(1,k+1):
            sin += gcd(m,tmp)
print(sin)

from math import gcd
key = int(input())
sin = 0
for i in range(1,key+1):
    for j in range(1,key+1):
        temp = gcd(i,j)
        for m in range(1,key+1):
            sin += gcd(m,temp)
print(sin)

Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
def gcd(a,b):
  if b == 0:
    return a
  else:
    return gcd(b,a%b)


k=input()
ans=0
for i in range(1,k+1):
  for j in range(1,k+1):
    d=gcd(i,j)
    for l in range(1,k+1):
      if d!=1:
        ans+=gcd(d,l)
      else:
        ans+=1


print(ans)

import sys
sys.setrecursionlimit(4100000)


def gcd(a,b):
  if b == 0:
    return a
  else:
    return gcd(b,a%b)


k=int(input())
ans=0
for i in range(1,k+1):
  for j in range(1,k+1):
    d=gcd(i,j)
    for l in range(1,k+1):
      if d!=1:
        ans+=gcd(d,l)
      else:
        ans+=1


print(ans)

Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math
from functools import reduce

def gcd(*numbers):
    return reduce(math.gcd, numbers)

N = int(input())
i = 0

for a in range(1, N+1):
    for b in range(1, N+1):
        for c in range(1, N+1):
            i += gcd(a, b, c)


from math import gcd 
N = int(input())
i = 0
 
for a in range(1, N+1):
    for b in range(1, N+1):
        for c in range(1, N+1):
            i += gcd(gcd(a, b), c)
            
print(i)

Verdicts: Wrong Answer

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
k=int(input())
ans=0
from fractions import gcd
import functools 
for a in range(1,k+1):
    for b in range(1,k+1):
        for c in range(1,k+1):
            @functools.lru_cache(None) 
            def gc(x,y):
                if x>y:
                    x,y=y,x
                return gcd(x,y)

            ans+=gcd(a,gcd(b,c))
print(ans)



k=int(input())
ans=0
from fractions import gcd
for a in range(1,k+1):
    for b in range(1,k+1):
        t=gcd(a,b)
        for c in range(1,k+1):
            ans+=gcd(c,t)
print(ans)



k=int(input())
ans=0
from fractions import gcd
for a in range(1,k+1):
    for b in range(1,k+1):
        for c in range(1,k+1):
#            print(a,b,c)
            ans+=gcd(a,gcd(b,c))
print(ans)



k=int(input())
ans=0
from math import gcd
for a in range(1,k+1):
    for b in range(1,k+1):
        t=gcd(a,b)
        for c in range(1,k+1):
            ans+=gcd(c,t)
print(ans)



Verdicts: Time Limit Exceeded

Time Limit Exceeded

Time Limit Exceeded

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math

def gcd(*numbers):
    return reduce(math.gcd, numbers)

K = int(input())
sum = 0
for i in range(1, K+1):
  for j in range(1, K+1):
    for k in range(1, K+1):
      sum += gcd(i,j,k)
print(sum)

import math
K=int(input())
ans=0
for i in range(1,K+1):
    for j in range(1,K+1):
        tmp=math.gcd(i,j)
        for k in range(1,K+1):
            ans+=math.gcd(tmp,k)         
print(ans)

Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import numpy as np
import math

num = int(input())
a = 1
sum = 0

def gcd(a, b, c):
  num = np.array([a, b, c])
  return np.gcd(np.gcd.reduce(num))

def loopC(a, b, num):
  c = 1
  sum = 0
  while c <= num:
      ans = gcd(a, b, c)
      sum += ans
      c += 1
  return sum

def loopB(a, num):
  b = 1
  sum = 0
  while b <= num:
    ans = loopC(a, b, num)
    b += 1
    sum += ans
  return sum

while a <= num:
  ans = loopB(a, num)
  sum += ans
  a += 1

print(sum)

import math

num = int(input())
sum = 0

for a in range(1, num + 1):
  for b in range(1, num + 1):
    gcd_ab = math.gcd(a, b)
    for c in range(1, num + 1):
      sum += math.gcd(gcd_ab, c)

print(sum)

Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math
k = int(input())

ans = 0
for a in range(k+1, 1, -1):
    for b in range(k+1, 1, -1):
        for c in range(k+1, 1, -1):
            ans += math.gcd(a, math.gcd(b, c))
print(ans)

import math

k = int(input())

ans = 0
for a in range(1, k+1):
    for b in range(1, k+1):
        tmp = math.gcd(a, b)
        for c in range(1, k+1):
            ans += math.gcd(tmp, c)
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
N = int(input())
s = 0
for i in range(0,N):
  if (i+1) % 3 1= 0 or (i+1) % 5 != 0:
    s = s + (i+1)
print(s)

K = int(input())
S = 0

def gcd(a,b):
   while b != 0:
       a, b = b, a % b
       if b ==0:
           return a
A1 = [0,0]
A2 = [0,0,0]
for i in range(1, K):
    S = S + i

if K > 1:
    for j in range(1, K):
        A1[0] =  j
        for k in range(j+1,K+1):
            A1[1] = k
            S += 6 * gcd(A1[1],A1[0])

if K > 2:
    for i1 in range(1, K-1):
        A2[0] = i1
        for k1 in range(i1+1, K):
            A2[1] = k1
            for j1 in range(k1+1,K+1):
                A2[2] = j1
                G3 = gcd(A2[2],A2[1])
                if G3 >= A2[0]:
                    S += 6 * gcd(G3,A2[0])
                else:
                    S += 6 * gcd(A2[0],G3)
                    
print(S)

K = int(input())
S = 0

def gcd(a,b):
   while b != 0:
       a, b = b, a % b
       if b ==0:
           return a
A1 = [0,0]
A2 = [0,0,0]
for i in range(1, K+1):
    S = S + i

if K > 1:
    for j in range(1, K):
        A1[0] =  j
        for k in range(j+1,K+1):
            A1[1] = k
            S += 6 * gcd(A1[1],A1[0])

if K > 2:
    for i1 in range(1, K-1):
        A2[0] = i1
        for k1 in range(i1+1, K):
            A2[1] = k1
            for j1 in range(k1+1,K+1):
                A2[2] = j1
                G3 = gcd(A2[2],A2[1])
                if G3 >= A2[0]:
                    S += 6 * gcd(G3,A2[0])
                else:
                    S += 6 * gcd(A2[0],G3)
print(S)

Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math
from functools import reduce

k = int(input())
ans = 0

def gcd(numbers):
    return reduce(math.gcd, numbers)

for i in range(k):
  for j in range(k):
    for k in range(k):
      ans += gcd([i+1, j+1, k+1])

print(ans)

from math import gcd

def solve():
    K = int(input())+1
    ans = 0
    for i in range(1,K):
        for j in range(1,K):
            for k in range(1,K):
                ans += gcd(gcd(i, j),k)
    return ans

print(solve())

Verdicts: Wrong Answer

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import numpy as np
import math
K = int(input())
memo = np.zeros((K,K),int)
s = 0
for i in range(K):
  for j in range(i,K+1):
    for k in range(j,K+1):
      if memo[i][j] != 0:
        if memo[memo[i][j]-1][k] != 0:
          s += memo[memo[i][j]-1][k]
        else:
          memo[memo[i][j]-1][k] = math.gcd(memo[i][j],k+1)
          memo[k][memo[i][j]-1] = memo[memo[i][j]-1][k]
          s += memo[k][memo[i][j]-1]
      else:
        
        memo[i][j] = math.gcd(i+1,j+1)
        memo[j][i] = memo[i][j]
        if memo[memo[i][j]-1][k] != 0:
          s += memo[memo[i][j]-1][k]
        else:
          memo[memo[i][j]-1][k] = math.gcd(memo[i][j],k+1)
          memo[k][memo[i][j]-1] = memo[memo[i][j]-1][k]
          s += memo[k][memo[i][j]-1]
print(s)


import numpy as np
import math
K = int(input())
memo = np.zeros((K,K),int)
s = 0
for i in range(K):
  for j in range(i,K):
    for k in range(j,K):
      if i == j and j == k:
        rate = 1
      elif i == j or j == k or i == k:
        rate = 3
      else:
        rate = 6
      s += math.gcd(math.gcd(i+1,j+1),k+1) * rate
print(s)


Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
import math
from functools import reduce

def gcd(*numbers):
    return reduce(math.gcd, numbers)

def gcd_list(numbers):
    return reduce(math.gcd, numbers)
  
k = int(input())
s = 0
for i in range(k):
  for j in range(k):
    for h in range(k):
      s += gcd(i,j,h)
print(s)

import math
from functools import reduce

def gcd(*numbers):
    return reduce(math.gcd, numbers)

def gcd_list(numbers):
    return reduce(math.gcd, numbers)
  
n,k = map(int, input().split())
s = 0
for i in range(k):
  for j in range(k):
    for h in range(k):
      s += gcd(i+1,j+1,h+1)
print(s)


import math
from functools import reduce

def gcd(*numbers):
    return reduce(math.gcd, numbers)

def gcd_list(numbers):
    return reduce(math.gcd, numbers)
  
k = int(input())
s = 0
l = []
for i in range(k):
  for j in range(k):
    l.append(math.gcd(i+1,j+1))
for h in range(k):
  for n in l:
    s += math.gcd(h+1, n)
print(s)


Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
k=int(input())
    ans=0
    for x in range(1,k+1):
        for y in range(1,k+1):
            for z in range(1,k+1):
               if math.gcd(x,y)==1:
                   ans+=1
               
               elif math.gcd(math.gcd(x,y),z)==1: 
                   ans+=1
               else: 
                       ans+=math.gcd(math.gcd(x,y),z)
    print(ans)

import numpy as np
K=int(input())
x = np.arange(1, K + 1)
nums = np.gcd.outer(np.gcd.outer(x, x), x)
print(nums.sum())

Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
K = int(input())
def gcd(x, y):
  a, b = x, y
  while b != 0:
    a, b = b, a % b
  return a
gcds = [[0 for j in range(K)] for i in range(K)]
for i in range(K):
  for j in range(K):
    gcds[i][j] = gcd(i+1, j+1)
S = 0
for i in range(K):
  for j in range(K):
    for k in range(K):
      S += gcds[gcds[i][j]][k]
print(S)

K = int(input())
mod = int(1e9) + 7
X = [0 for _ in range(K)]
def doubling(n, m):
  y = 1
  base = n
  while m != 0:
    if m % 2 == 1:
      y *= base
      y %= mod
    base *= base
    base %= mod
    m //= 2
  return y
SS = 0
for i in range(K, 0, -1):
  d = K // i
  S = doubling(d, 3)
  if d > 1:
    for j in range(2, d+1):
      S -= X[j*i-1]
      S %= mod
  X[i-1] = S
  SS += i * S
  SS %= mod
print(SS)

Verdicts: Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
from math import gcd
N = int(input())
sum = 0
for i in range(1, N+1):
    for j in range(1, N+1):
		tmp=gcd(j,k);
        for k in range(1, N+1):
            sum += gcd(i, tmp)
print(sum)

from math import gcd
N = int(input())
sum = 0
for i in range(1, N+1):
    for j in range(1, N+1):
		tmp=gcd(j,k)
        for k in range(1, N+1):
            sum += gcd(i, tmp)
print(sum)


from math import gcd
N = int(input())
sum = 0
for i in range(1, N+1):
    for j in range(1, N+1):
        tmp=gcd(i,j)
        for k in range(1, N+1):
            sum += gcd(tmp,k)
print(sum)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: Find \displaystyle{\sum_{a=1}^{K}\sum_{b=1}^{K}\sum_{c=1}^{K} \gcd(a,b,c)}.  Here \gcd(a,b,c) denotes the greatest common divisor of a, b, and c.

Submission attempts (in order):
k = int(input())

#acd_list = []
all_num = [i for i in range(1,k + 1)]
ans = 0 

#comb_list  = list(itertools.combinations(all_num, 3))
#print(comb_list)

def gcd(*numbers):
    return reduce(math.gcd, numbers)

def gcd_list(numbers):
    return reduce(math.gcd, numbers)

result = []
for i in all_num:
  for j in all_num:
    for k in all_num:
      result.append([i, j, k])

for a,b,c in result:
#for a,b,c in comb_list:
    s = gcd(a, b,c)
    ans += s
print(ans)

import math
 
k = int(input()) + 1
ans = 0
 
for a in range(1, k):
    for b in range(1, k):
        d = math.gcd(a, b)
        for c in range(1, k):
            ans += math.gcd(d, c)
 
print(ans)

Verdicts: Runtime Error

Accepted
Problem: We have a string S of length N consisting of `R`, `G`, and `B`.  Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:    * S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.   * j - i \neq k - j.

Submission attempts (in order):
n=int(input())
s=list(input())
ans=s.count('R')*s.count('G')*s.count('B')
for i in range(n):
  if s[i]=='R':
    s[i]=1
    elif s[i]=='G':
      s[i]=2
    elif s[i]=='B':
      s[i]=4
for i in range(n):
   for j in range(i+1,n):
      k=2*j-i
      if k>=n:
        continue
      if s[i]+s[j]+s[k]==7:
        ans-=1
print(ans)


n=int(input())
s=list(input())
ans=s.count('R')*s.count('G')*s.count('B')
for i in range(n):
  if s[i]=='R':
    s[i]=1
    elif s[i]=='G':
      s[i]
    elif s[i]=='B':
      s[i]=4
for i in range(n):
   for j in range(i+1,n):
      k=2*j-i
      if k>=n:
        continue
      if s[i]+s[j]+s[k]==7:
        ans-=1
print(ans)

n=int(input())
s=list(input())
ans=s.count('R')*s.count('G')*s.count('B')
for i in range(n):
   if s[i]=='R':
       s[i]=1
   elif s[i]=='G':
       s[i]
   elif s[i]=='B':
       s[i]=4
for i in range(n):
   for j in range(i+1,n):
     k=2*j-i
     if k>=n:
       continue
     if s[i]+s[j]+s[k]==7:
         ans-=1
print(ans)

n=int(input())
s=list(input())
ans=s.count('R')*s.count('G')*s.count('B')
for i in range(n):
  if s[i]=='R':
    s[i]=1
  elif s[i]=='G':
    s[i]=2
  elif s[i]=='B':
    s[i]=4
for i in range(n):
   for j in range(i+1,n):
      k=2*j-i
      if k>=n:
        continue
      if s[i]+s[j]+s[k]==7:
        ans-=1
print(ans)


Verdicts: Runtime Error

Runtime Error

Runtime Error

Accepted
Problem: We have a string S of length N consisting of `R`, `G`, and `B`.  Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:    * S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.   * j - i \neq k - j.

Submission attempts (in order):
N = int(input())
R = []
G = []
B = []
s =input()
ans = 0
for i in range(N):
    if s[i] == "R":
        R.append(i)
    elif s[i] == "G":
        G.append(i)
    else:
        B.append(i)
sB = set(B)
print(sB)
ans = len(R) * len(G) * len(B)
for r in R:
    for g in G:
        i = min(r,g)
        j = max(r,g)
        sa = j - i
        if i - sa in sB:
            ans -= 1
        if j + sa in sB:
            ans -= 1
        if (i + j) / 2 in sB:
            ans -= 1
print(ans)

N = int(input())
R = []
G = []
B = []
s =input()
ans = 0
for i in range(N):
    if s[i] == "R":
        R.append(i)
    elif s[i] == "G":
        G.append(i)
    else:
        B.append(i)
sB = set(B)
ans = len(R) * len(G) * len(B)
for r in R:
    for g in G:
        i = min(r,g)
        j = max(r,g)
        sa = j - i
        if i - sa in sB:
            ans -= 1
        if j + sa in sB:
            ans -= 1
        if (i + j) / 2 in sB:
            ans -= 1
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: We have a string S of length N consisting of `R`, `G`, and `B`.  Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:    * S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.   * j - i \neq k - j.

Submission attempts (in order):
n = int(input())
s = input()

r = [0] * n
g = [0] * n
b = [0] * n
dc = {'R':[0]*n,'G':[0]*n,'B':[0]*n}
for i in range(n-1):
    dc['R'][i] = dc['R'][i + 1]
    dc['G'][i] = dc['G'][i + 1]
    dc['B'][i] = dc['B'][i + 1]
    dc[s[i]] += 1
# print(r)
# print(g)
# print(b)
dcc = {('R','G'):'B',('R','B'):'G',('B','G'):'R',('G','R'):'B',('B','R'):'G',('G','B'):'R'}
if(n <= 3):print(0)
else:
    c = 0
    for i in range(n):
        for j in range(i+1,n):
            if(s[i] == s[j]):continue
            d = j - i
            c += dc[dcc[(s[i],s[j])]][j]
            k = j + d
            if(k < n):
                if(s[k] == dcc[(s[i],s[j])]):c -= 1
print(c)

n = int(input())
s = input()
d = [0,0,0]
for i in s:
    if(i == 'R'):d[0] += 1
    elif(i == 'G'):d[1] += 1
    elif(i == 'B'):d[2] += 1
c = d[0] * d[1] * d[2]

for i in range(n):
    for j in range(i+1,n):
        k = 2 * j - i
        if(k >= n):continue
        if(s[i] == s[j]):continue
        if(s[j] == s[k]):continue
        if(s[i] == s[k]):continue
        c -= 1

print(c)

Verdicts: Runtime Error

Accepted
Problem: We have a string S of length N consisting of `R`, `G`, and `B`.  Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:    * S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.   * j - i \neq k - j.

Submission attempts (in order):
n=int(input())
s=input()
ans=0
r=[]
g=[]
b=[]
for i in range(n):
    tmp=s[i]
    if tmp=="R":
        r.append(i+1)
    elif tmp=="G":
        g.append(i+1)
    else:
        b.append(i+1)
for i in r:
    for j in g:
        for k in b:
            if (i<j and j<k and i<k) or (i<k and k<j and i<j):
                continue
            if j-i==k-j:
                continue
            else:
                ans+=1
print(ans)

n=int(input())
s=input()
r=s.count("R")
g=s.count("G")
b=s.count("B")
ans=r*g*b
for i in range(n):
    for j in range(n):
        if i+j+j>=n:
            break
        s1=s[i]
        s2=s[i+j]
        s3=s[i+j*2]
        if s1!=s2 and s2!=s3 and s1!=s3:
            ans-=1
print(ans)

Verdicts: Wrong Answer

Accepted
Problem: We have a string S of length N consisting of `R`, `G`, and `B`.  Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:    * S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.   * j - i \neq k - j.

Submission attempts (in order):
N = int(input())
S = input()
Sr = Sg = Sb = 0
for c in S:
    if c == 'R':
        Sr += 1
    elif c == 'G':
        Sg += 1
    else:
        Sb += 1
A = Sr * Sg * Sb
for sep in range(1,(N-1)//2):
    for i in range(N-2*sep):
        if S[i] != S[i+sep] and S[i+sep] != S[i+2*sep] and S[i] != S[i+2*sep]:
            A -= 1
print(A)


N = int(input())
S = input()
Sr = Sg = Sb = 0
for c in S:
    if c == 'R':
        Sr += 1
    elif c == 'G':
        Sg += 1
    else:
        Sb += 1
A = Sr * Sg * Sb
for sep in range(1,(N+1)//2):
    for i in range(N-2*sep):
        if S[i] != S[i+sep] and S[i+sep] != S[i+2*sep] and S[i] != S[i+2*sep]:
            A -= 1
print(A)


Verdicts: Wrong Answer

Accepted
Problem: We have a string S of length N consisting of `R`, `G`, and `B`.  Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:    * S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.   * j - i \neq k - j.

Submission attempts (in order):
# -*- coding: utf-8 -*-

import sys
args = sys.argv

debug_flg = False
def debug(val1,val2="",val3="",val4=""):
    global debug_flg
    if debug_flg:
        print(val1,val2,val3,val4,sep=" , ")

if len(args) != 2:
    n= int(input())
    s= input()
else:
    debug_flg = True
    if args[1] == "2":
        n=39
        s="RBRBGRBGGBBRRGBBRRRBGGBRBGBRBGBRBBBGBBB"
    else:
        n=4
        s="RRGB"

snum = []
for i in range(0,n):
    snum.append(s[i-1])

ret =0
for i in range(1,n-1):
    for j in range(i,n):
        if snum[i-1] == snum[j-1]:
            continue
        for k in range(j,n+1):
            if j-i == k-j:
                continue
            if snum[i-1] == snum[k-1]:
                continue
            if snum[j-1] == snum[k-1]:
                continue
            ret += 1

print(ret)






# -*- coding: utf-8 -*-

import sys
args = sys.argv

debug_flg = False
def debug(val1,val2="",val3="",val4=""):
    global debug_flg
    if debug_flg:
        print(val1,val2,val3,val4,sep=" , ")

if len(args) != 2:
    n= int(input())
    s= input()
else:
    debug_flg = True
    if args[1] == "2":
        n=39
        s="RBRBGRBGGBBRRGBBRRRBGGBRBGBRBGBRBBBGBBB"
    else:
        n=4
        s="RRGB"

snum = []
for i in range(0,n):
    if s[i-1] == "R":
        snum.append(0)
    elif s[i-1] == "G":
        snum.append(1)
    elif s[i-1] == "B":
        snum.append(2)

ret =0
for i in range(1,n-1):
    for j in range(i,n):
        if snum[i-1] == snum[j-1]:
            continue
        for k in range(j,n+1):
            if j-i == k-j:
                continue
            if snum[i-1] == snum[k-1]:
                continue
            if snum[j-1] == snum[k-1]:
                continue
            ret += 1

print(ret)

def main():
    n= int(input())
    s= input()

    r = s.count("R")
    g = s.count("G")
    b = s.count("B")

    ans = r * g * b
    snum = tuple(s[:])
    for i in range(1,n-1):
        for j in range(i,n):
            k = j * 2 - i
            if k > n:
                break

            if snum[i-1] != snum[j-1] and snum[i-1] != snum[k-1] and snum[j-1] != snum[k-1]:
                ans -= 1

    print(ans)


if __name__ == '__main__':
    main()



Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: We have a string S of length N consisting of `R`, `G`, and `B`.  Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:    * S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.   * j - i \neq k - j.

Submission attempts (in order):
import collections
n = int(input())
a = list(input())
b = collections.Counter(a)
ans =b['R']*b['G']*b['B']
print(b,ans)
for i in range(n-2):
    for j in range(i+1,n-1):
        if a[i] == a[j]:
            continue
        c = j*2 -i
        if c < n:
            if a[i] != a[j] and a[i] != a[c] and a[j] != a[c]:
                ans -= 1
print(ans)

import sys
input = sys.stdin.readline
n = int(input())
s = input()
s = s.replace('R','0')
s = s.replace('G','1')
s = s.replace('B','2')
s = tuple(map(int,s))
rgb = [[0 for _ in range(3)] for _ in range(n+1)]
for i in range(n):
    for j in range(3):
        rgb[i+1][j] = rgb[i][j]
    rgb[i+1][s[i]] = rgb[i][s[i]] +1
rgb = tuple(rgb)
ans = 0
for i in range(n):
    a = s[i]
    for j in range(i+1,n):
        b = s[j]
        if a != b:
            c = 3 -a- b
            ans += rgb[n][c] - rgb[j+1][c]
            if 2*j-i < n and s[2*j-i] == c:
                ans -= 1

print(ans)


import collections
n = int(input())
a = input()
a = a.replace('R','1')
a = a.replace('G','2')
a = a.replace('B','3')
a = tuple(map(int,list(a)))
b = collections.Counter(a)
ans =b[3]*b[1]*b[2]
for i in range(n-2):
    for j in range(i+1,n):
        if a[i] == a[j]:
            continue
        c = j*2 -i
        if c < n:
            if a[i]+a[j]+a[c]==6:
                ans -= 1
        else:
            break
print(ans)

Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: We have a string S of length N consisting of `R`, `G`, and `B`.  Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:    * S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.   * j - i \neq k - j.

Submission attempts (in order):
from sys import stdin
n = int(input())
s = stdin.readline()
rind = []
gind = []
bind = []
ans = 0
for i in range(n):
    if s[i] == "R":
        rind.append(i+1)
    elif s[i] == "G":
        gind.append(i+1)
    else:
        bind.append(i+1)
print(rind,gind,bind)
for i in range(len(rind)):
    for j in range(len(gind)):
        for k in range(len(bind)):
            ind = [rind[i],gind[j],bind[k]]
            a = sorted(ind)
            if a[0] + a[2] != 2*a[1]:
                ans += 1
print(ans)


from sys import stdin
n = int(input())
s = stdin.readline()
r = 0
g = 0 
b = 0
for i in range(n):
    if s[i] == "R":
        r += 1
    elif s[i] == "G":
        g += 1
    else:
        b += 1
count1 = r*g*b
count2 = 0
for i in range(0,n-3):
    for j in range(i+1,n-2):
        if 2*j-i<=n-1:
            if (s[i] != s[j])and(s[i] != s[2*j-i])and(s[j] != s[2*j-i]):
               count2 += 1
print(count1-count2)

from sys import stdin
n = int(input())
s = stdin.readline()
rind = []
gind = []
bind = []
ans = 0
for i in range(n):
    if s[i] == "R":
        rind.append(i+1)
    elif s[i] == "G":
        gind.append(i+1)
    else:
        bind.append(i+1)
for i in range(len(rind)):
    for j in range(len(gind)):
        for k in range(len(bind)):
            if (rind[i]+gind[j]!=2*bind[k])and(gind[j]+bind[k]!=rind[i])and(bind[k]+rind[i]!=gin[j]):
                ans += 1
print(ans)


from sys import stdin
n = int(input())
s = stdin.readline()
r = 0
g = 0 
b = 0
for i in range(n):
    if s[i] == "R":
        r += 1
    elif s[i] == "G":
        g += 1
    else:
        b += 1
count1 = r*g*b
count2 = 0
for i in range(0,n-2):
    for j in range(i+1,n-1):
        if 2*j-i<=n-1:
            if (s[i] != s[j])and(s[i] != s[2*j-i])and(s[j] != s[2*j-i]):
               count2 += 1
print(count1-count2)

Verdicts: Wrong Answer

Wrong Answer

Runtime Error

Accepted
Problem: We have a string S of length N consisting of `R`, `G`, and `B`.  Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:    * S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.   * j - i \neq k - j.

Submission attempts (in order):
import itertools
from collections import Counter

N = int(input())
S = input()

C=Counter(S)
r = C['R']
g = C['G']
b = C['B']


cnt = r * g * b

for i in range(0,N-2):
    for j in range(i,N-1):
        k = j + j - i
        if k < N:
            if S[i] != S[j] and S[i] != S[k] and S[j] != S[k]):
                cnt -= 1

print(cnt)

import itertools
from collections import Counter

N = int(input())
S = input()

C=Counter(S)
r = C['R']
g = C['G']
b = C['B']


cnt = r * g * b

sum = 0
for i in range(0,N-2):
    for j in range(i,N-1):
        k = j + j - i
        if k < N
            if (S[i] != S[j]) and (S[i] != S[k] and S[j] != S[k]):
                cnt -= 1

print(cnt)


from collections import Counter

def SUMCHK():
    N = int(input())
    S = input()

    C=Counter(S)
    r = C['R']
    g = C['G']
    b = C['B']


    cnt = r * g * b

    for i in range(0,N-2):
        for j in range(i,N-1):
            k = j + j - i
            if k < N:
                if S[i] != S[j] and S[i] != S[k] and S[j] != S[k]:
                    cnt -= 1

    print(cnt)

SUMCHK()


Verdicts: Runtime Error

Runtime Error

Accepted
Problem: We have a string S of length N consisting of `R`, `G`, and `B`.  Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:    * S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.   * j - i \neq k - j.

Submission attempts (in order):
n = int(input())
s = input()
ans = 0
ans_list = []

R_index = []
G_index = []
B_index = []

for i in range(n):
    if s[i] == 'R':
        R_index.append(i)
    elif s[i] == 'G':
        G_index.append(i)
    elif s[i] == 'B':
        B_index.append(i)

ans = len(R_index) * len(G_index) * len(B_index)

for j in range(1, n-1):
    for i in range(min(n-j-1,j)):
        if s[j-i] != s[j] and s[j] != s[j+i] and s[j-i] != s[j+i]:
            ans -= 1


print(ans)


n = int(input())
s = input()
ans = 0
ans_list = []

R_index = []
G_index = []
B_index = []

for i in range(n):
    if s[i] == 'R':
        R_index.append(i)
    elif s[i] == 'G':
        G_index.append(i)
    elif s[i] == 'B':
        B_index.append(i)

ans = len(R_index) * len(G_index) * len(B_index)

for j in range(1, n-1):
    for i in range(min(n-j-1,j)+1):
        # print(j,i)
        # print(min(n-j-1,j))
        #
        # print(s[j-i],s[j],s[j+i])
        if s[j-i] != s[j] and s[j] != s[j+i] and s[j-i] != s[j+i]:
            # print('here')
            ans -= 1

print(ans)


Verdicts: Wrong Answer

Accepted
Problem: We have a string S of length N consisting of `R`, `G`, and `B`.  Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:    * S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.   * j - i \neq k - j.

Submission attempts (in order):
import sys
import collections


def main():
    input = sys.stdin.readline
    N = int(input())
    S = input()
    A = S.count('R')*S.count('G')*S.count('B')
    total = 0
    for j in range(1, N-1):
        for i in range(j):
            k = 2*j-i
            order += 1
            if k < len(S):
                if S[i] != S[j] and S[i] != S[k] and S[j] != S[k]:
                    total += 1
    print(A - total)


def main2():
    N = int(input())
    S = input()
    A = S.count('R')*S.count('G')*S.count('B')
    order = 0
    for i in range(N-2):
        for j in range(i+1, N-1):
            order += 1
            if 2*j-i < N:
                if S[i] != S[j] and S[i] != S[2*j-i] and S[j] != S[2*j-i]:
                    A -= 1
    print(A)
    print(order)


if __name__ == "__main__":
    main()


import sys


def main():
    input = sys.stdin.readline
    N = int(input())
    S = input().strip()
    A = S.count('R')*S.count('G')*S.count('B')
    assert 1 <= N <= 4000
    assert N == len(S)
    total = 0
    for j in range(1, N-1):
        for i in range(j):
            k = 2*j-i
            if k < len(S):
                if S[i] != S[j] and S[i] != S[k] and S[j] != S[k]:
                    total += 1
    print(A - total)


if __name__ == "__main__":
    main()


Verdicts: Runtime Error

Accepted
Problem: We have a string S of length N consisting of `R`, `G`, and `B`.  Find the number of triples (i,~j,~k)~(1 \leq i < j < k \leq N) that satisfy both of the following conditions:    * S_i \neq S_j, S_i \neq S_k, and S_j \neq S_k.   * j - i \neq k - j.

Submission attempts (in order):
n = int(input())
s = input()
r = s.count("R");g = s.count("G");b = s.count("B")
count = r*g*b
print(count)
for j in range(n):
  for i in range(j):
    if (k:=2*j-i)<n:
      if s[i]!=s[k]!=s[j]!=s[i]:
        count-=1   
print(count)

n = int(input())
s = input()
r,g,b = 0,0,0
r = s.count("R");g = s.count("G");b = s.count("B")
count = r*g*b
print(count)
for j in range(n):
  for i in range(j):
    if (k:=2*j-i)<n:
      if s[i]!=s[k]!=s[j]!=s[i]:
        count-=1   
print(count)

n = int(input())
s = input()
r,g,b = 0,0,0
r = s.count("R");g = s.count("G");b = s.count("B")
k=0    
count = r*g*b
print(count)
for j in range(n):
  for i in range(j):
    if (k==2*j-i)<n:
      if s[i]!=s[k]!=s[j]!=s[i]:
        count-=1   
print(count)

n = int(input())
s = input()
r = s.count("R");g = s.count("G");b = s.count("B")
count = r*g*b
for j in range(n):
  for i in range(j):
    if (k:=2*j-i)<n:
      if s[i]!=s[k]!=s[j]!=s[i]:
        count-=1   
print(count)


Verdicts: Wrong Answer

Wrong Answer

Wrong Answer

Accepted