SGU - 117 - Counting （快速幂取模！）

SGU - 117

Counting

Time Limit: 250MS

Memory Limit: 4096KB

64bit IO Format: %I64d & %I64u

Submit Status

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

Author	: Michael R. Mirzayanov
Resource	: PhTL #1 Training Contests
Date	: Fall 2001

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;  
}