Poj1995--Raising Modulo Numbers（快速幂）

题目: http://poj.org/problem?id=1995

Input

The input consists of Z assignments. The number of them is given by the single positive integer Z appearing on the first line of input. Then the assignements follow. Each assignement begins with line containing an integer M (1 <= M <= 45000). The sum will be divided by this number. Next line contains number of players H (1 <= H <= 45000). Next exactly H lines follow. On each line, there are exactly two numbers Ai and Bi separated by space. Both numbers cannot be equal zero at the same time.

Output

For each assingnement there is the only one line of output. On this line, there is a number, the result of expression

(A₁^B₁+A₂^B₂+ ... +A_H^B_H)mod M.

快速幂，裸题：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
long long Deal(long long a, long long b, int mod)
{
    long long ans = 1;
    while(b)
    {
        if(b & 1)
            ans = ans*a%mod;
        b /= 2;
        a = (a*a)%mod;    
    }    
    return ans;
} 
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int mod, h;
        long long a, b, sum = 0;
        scanf("%d %d", &mod, &h);
        for(int i = 0; i < h; i++)
        {
            scanf("%lld %lld", &a, &b);
            sum += Deal(a, b, mod) % mod; 
        }
        printf("%lld
", sum%mod);
    }
    return 0;     
}