如何使用模 10^9+7

Question

I am trying to write a code for sum of square of natural numbers but with mod it's giving wrong answer. What would be the correct way here?

#include <bits/stdc++.h>
using namespace std;
#define mod 1000000007

int main()
{
    int N;
    cin>>N;
    cout<< (((N) * (N+1) * (2*N+1))/6)%mod;

    return 0;
}

Answer 1

(N) * (N+1) * (2*N+1) can be, even if N is less than 1000000007, too large. Namely up to 2000000039000000253000000546, which is an 91-bit number. It is not likely that int on your system is capable of containing such large numbers.

As usual with this type of question, the work-around is a combination of:

• Using a larger integer type for intermediate products, and
• Applying modulo reduction on intermediate products, not only at the end.

Using just one of them is not sufficient.

As a consequence of applying modulo reduction earlier, the division by 6 will not work with a normal division, it will have to be a multiplication by the modular multiplicative inverse of 6 mod 1000000007, which is 166666668.

Example code:

mod_mul(mod_mul(mod_mul(N, N + 1, mod), mod_mul(2, N, mod) + 1, mod), inv6, mod)

Using some suitable definition of mod_mul , which can avoid overflow by using long long or a similar type.