SGU117—Counting （快速幂取模）_counting testcsdn-优快云博客

本文链接：https://blog.youkuaiyun.com/bingsanchun1/article/details/17011603

本文探讨了一个特定的编程挑战，即计算给定序列中多少个数的M次幂可以被K整除。介绍了使用快速幂算法解决该问题的方法，并提供了完整的C++实现代码。

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117. Counting

time limit per test: 0.5 sec.
memory limit per test: 4096 KB

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

题目是说，样例的第一行为三个数n，m，k；然后第二行有n个数，求第二行中，每个数的m次方能被k整除的个数。——快速幂

//题目大意：给定n个数，问这n个数的m次幂有多少能够被K整除。
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<ctype.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<math.h>
#include<vector>
#include<map>
#include<deque>
#include<list>
using namespace std;
int a[99999];
int quick_pow(int n,int m,int k)
{
    int result=1;
    while(m)
    {
        if(m&1)
        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 count=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            if(quick_pow(a[i],m,k)==0)
            count++;
        }
        printf("%d\n",count);
    }
    return 0;
}