求 (1! 2! 3! 4! ...... n!) % 494273703.

Question

#include<stdio.h>
#define ll long long int
#define P 494273703

int main()
{
    ll n;
    scanf("%lld", &n);
    ll sum = 1, res = 0;
    ll i;
    for (i = 1; i <= n; i++)
    {
        sum *= i;
        sum %= P;
        res += sum;
        res %= P;
    }
    printf("%lld", res);
    return 0;
}

这样就会超时，因该怎样做

Answer 1

求余运算效率比较低，可以在求余前先判断值是否大于P，这样小于P就不用求余了