print(int(input())%7)

in python: -5 % 7 == 2
in c++: -5 % 7 == -5

print(int(input()) % 7)

a = int(input())
c= a//7
d=c*7
print(a-d)

I got my code running due to the solutions posted.. But i want to understand the logic behind the division.. Any one?

#include<bits/stdc++.h>

using namespace std;

int main(){
    unsigned long long int n;
    try{
        cin >> n;
        cout << n%7 << endl;
    }catch(exception e){
        cout << 1 << endl;
    }
    return 0;
}

#include <stdio.h>
int main() {
unsigned long long int a;
scanf("%llu", &a);
if (a/7!=0)
printf("%llu\n", a%7);
else if (a < 7)
printf("%d\n", a); }

https://umath.ru/theory/priznak-delimosti-na-7/

s = 0
k = 1
x = input()
while len(x) > 3:
    s += int(x[-3:]) * k
    k *= -1
    x = x[:-3]
s += int(x) * k
print(s % 7)

Edited by author 25.01.2019 02:16

#include <cstdio>
char ch, res;
int main (){
    while(scanf(" %c", &ch)!= EOF||(printf("%d\n", res)&0))
        res= (res*10+ ch-'0')%7;
}

#include <iostream>
using namespace std;


int main()
{
    int k = 0;
    long long int s = 0;
    char num[52];
    int arr[51];
    bool flag = true;
    cin >> num;
    for(int i = 0; flag; i++)
    {
        switch(num[i])
        {
            case '0':
            {
                arr[i] = 0;
                k++;
                break;
            }
            case '1':
            {
                arr[i] = 1;
                k++;
                break;
            }
            case '2':
            {
                arr[i] = 2;
                k++;
                break;
            }
            case '3':
            {
                arr[i] = 3;
                k++;
                break;
            }
            case '4':
            {
                arr[i] = 4;
                k++;
                break;
            }
            case '5':
            {
                arr[i] = 5;
                k++;
                break;
            }
            case '6':
            {
                arr[i] = 6;
                k++;
                break;
            }
            case '7':
            {
                arr[i] = 7;
                k++;
                break;
            }
            case '8':
            {
                arr[i] = 8;
                k++;
                break;
            }
            case '9':
            {
                arr[i] = 9;
                k++;
                break;
            }
            case '\0':
            {
                flag = false;
                break;
            }
        }
    }
    for(int i = k-1; i+1 > 6;)
    {
        s = s + arr[i];
        s = s + arr[i-1]*3;
        s = s + arr[i-2]*2;
        s = s + arr[i-3]*6;
        s = s + arr[i-4]*4;
        s = s + arr[i-5]*5;
        i = i - 6;
        k = k - 6;
    }
    if(k > 0)
    {
        s = s + arr[k-1];
        k--;
    }
    if(k > 0)
    {
        s = s + arr[k-1]*3;
        k--;
    }
    if(k > 0)
    {
        s = s + arr[k-1]*2;
        k--;
    }
    if(k > 0)
    {
        s = s + arr[k-1]*6;
        k--;
    }
    if(k > 0)
    {
        s = s + arr[k-1]*4;
        k--;
    }
    if(k > 0)
    {
        s = s + arr[k-1]*5;
        k--;
    }
    cout << s%7;
    return 0;
}

Edited by author 16.11.2013 14:57

Edited by author 16.11.2013 14:57

компилятор visual studio 2010 - 1wa
c++11                          -4wa
c++14                          -4wa
КАК???!!!

#include <iostream>
#include <math.h>

using namespace std;

int main(){

    double x;
    int i;

    cin >> x;

    x = fmod (x, 7.0);

    printf("%.0f",x);

    system("pause");
    return 0;

}

Edited by author 18.08.2014 21:49

Just use it to solve this problem.

del

Edited by author 06.01.2015 21:37

var a:array[1..20] of integer;
n,n1:string;
x,k,i,s,code:integer;
begin
read(n);
x:=length(n);
k:=x div 3;
if x mod 3>0 then k:=k+1;
for i:=1 to k do begin
n1:=copy(n,x-3*i+1,3);
val(n1,k,code);
a[i]:=k;
delete(n,x-2*i,3);
end;
k:=x div 3;
if x mod 3>0 then k:=k+1;
if k div 2=0 then begin
a[1]:=-1*a[1];
a[3]:=-1*a[3];
a[5]:=-1*a[5];
a[7]:=-1*a[7];
a[9]:=-1*a[9];
a[11]:=-1*a[11];
a[13]:=-1*a[13];
a[15]:=-1*a[15];
a[17]:=-1*a[17];
end
else begin
a[2]:=-1*a[2];
a[4]:=-1*a[4];
a[6]:=-1*a[6];
a[8]:=-1*a[8];
a[10]:=-1*a[10];
a[12]:=-1*a[12];
a[14]:=-1*a[14];
a[16]:=-1*a[16];
end;
for i:=1 to 17 do
s:=s+a[i];
write(abs(s mod 7));
end.

what's that means?
Output limit exceeded.

Dear Java Coders, you will solve this problem so easy, if you know what is BigInteger means.

If you use BigInteger the program is e-a-s-y :) Got AC from the first try :)