How to check for division by 7 for big number in C++?

Question

I have to check, if given number is divisible by 7, which is usualy done just by doing something like n % 7 == 0, but the problem is, that given number can have

Answer 1

you can use a known rule about division by 7 that says: group each 3 digits together starting from the right and start subtracting and adding them alternativly, the divisibility of the result by 7 is the same as the original number:

ex.:

testing 341234612736481253489125349812643761283458123548213541273468213
        549821354182354891623458917245921834593218645921580

   (580-921+645-218+593-834+921-245+917-458+623-891+354-182
    +354-821+549-213+468-273+541-213+548-123+458-283+761-643
    +812-349+125-489+253-481+736-612+234-341 
    = 1882 )
    % 7 != 0 --> NOK!

there are other alternatives to this rule, all easy to implement.

Answer 2

You can compute the value of the number modulo 7.

That is, for each digit d and value n so far compute n = (10 * n + d) % 7.

This has the advantage of working independently of the divisor 7 or the base 10.

I solved this problem exactly the same way on one of programming contests. Here is the fragment of code you need:

int sum = 0;
while (true) {
  char ch;
  cin>>ch;
  if (ch<'0' || ch>'9') break; // Reached the end of stdin
  sum = sum*10; // The previous sum we had must be multiplied
  sum += (int) ch;
  sum -= (int) '0'; // Remove the code to get the value of the digit
  sum %= 7; 
}

if (sum==0) cout<<"1";
else cout<<"0";

This code is working thanks to simple rules of modular arithmetics. It also works not just for 7, but for any divisor actually.

Answer 3

At first Take That Big Number in string And then sum every digit of string. at last check if(sum%7==0)

Code:

#include <bits/stdc++.h>
using namespace std;
int main()
{
    long long int n,i,j,sum,k;
    sum=0;
    string s;
    cin>>s;
    for(i=0;i<s.length();i++)
    {
        sum=sum+(s[i]-'0');
    }

    if(sum%7==0)
    {
        printf("Yes\n");
    }
    else
    {
        printf("No\n");
    }

    return 0;
}