SGU - 117 - Counting （快速幂取模！）

SGU - 117

Counting

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Description

Find amount of numbers for given sequence of integer numbers such that after raising them to the M-th power they will be divided by K.

Input

Input consists of two lines. There are three integer numbers N, M, K(0<N, M, K<10001) on the first line. There are N positive integer numbers ? given sequence (each number is not more than 10001) ? on the second
line.

Output

Write answer for given task.

Sample Input

4 2 50
9 10 11 12

Sample Output

1

Source

题意：第一行先给出n,m,k三个整数，第二行给出n个整数，判断这些整数的m的次方是否能被k整除，，简单快速幂即可。。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;  

int q_pow(int n, int m, int k)
{
    int result=1;
    while(m)  //快速幂
    {
        if(m&1)     //判断是否为奇数
        result=(result*n)%k;
        n=n*n%k;
        m>>=1;
    }
    return result%k;
}  

int main()
{
    int n, m, k;
    while(scanf("%d %d %d", &n, &m, &k)!=EOF)
    {
        int ans = 0;
        for(int i=0; i<n; i++)
        {
        	int a;
            scanf("%d", &a);
            if(q_pow(a, m, k)==0)  ans++;
        }
        printf("%d\n", ans);
    }
    return 0;
}