标签：acm algorithm icpc sgu 快速幂取模
SGU - 117
| Time Limit: 250MS | Memory Limit: 4096KB | 64bit IO Format: %I64d & %I64u |
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;
}标签：acm algorithm icpc sgu 快速幂取模
原文地址：http://blog.csdn.net/u014355480/article/details/41911071
