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 2 509 10 11 12

Sample Output

1

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Author: Michael R. MirzayanovResource: PhTL #1 Training ContestsDate: 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;  }