Sgu-117-counting (Quick power modulo!) ）

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 integers numbers such that after raising them to the M-th Power they Would be divided by K.

Input

Input consists of lines. There is three integer numbers N, M, K (0<n, M, k<10001) on the first line. There is N positive integer numbers? Given sequence (each number isn't more than 10001)? On the second line.

Output

Write answer for given task.

Sample Input

4 2 509 10 11 12

Sample Output

1

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

Source

Test instructions: The first line gives n,m,k three integers, the second line gives n integers, judging whether the second of these integers can be divisible by K, simple and fast power can be.

AC Code:

#include <cstdio> #include <cstring> #include <algorithm>  using namespace std;  int Q_pow (int n, int m, int k)  {      int result=1;      while (m)  //fast power     {          if (m&1)     //Determines whether it is an odd         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;  }

Sgu-117-counting (Quick power modulo!) ）