快速切题 sgu117. Counting 分解质因数

117. Counting

time limit per test: 0.25 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

1
用时:30min

#include <cstdio>

#include <cstring>

#include <queue>

using namespace std;

const int maxnum=10001;

bool isntprime[maxnum];

int prime[maxnum],cnt;

int divide[maxnum][15],num[maxnum],len[maxnum][15];//divide[i][j]为分解i得到的第j个质数 len[i][j]为分解i得到的第j个质数在i中出现的次数 num[i]为分解i得到的质因数个数

int n,m,k;

void calc(){

    cnt=0;

    for(int i=2;i<maxnum;i++){

        if(!isntprime[i]){

            prime[cnt++]=i;

            divide[i][0]=i;

            num[i]=1;

            len[i][0]=1;

            for(int j=i+i;j<maxnum;j+=i){

                isntprime[j]=true;

                divide[j][num[j]]=i;

                int tmp=j;

                while(tmp%i==0){

                    tmp/=i;

                    len[j][num[j]]++;

                }

                num[j]++;

            }

        }

    }

}



int main(){

    calc();

    scanf("%d%d%d",&n,&m,&k);

    if(k==1){printf("%d",n);return 0;}

    int ans=0;

    for(int i=0;i<n;i++){

        int tmp;

        scanf("%d",&tmp);

        bool fl=true;

        for(int j=0;j<num[k];j++){

                int fnd=(lower_bound(divide[tmp],divide[tmp]+num[tmp],divide[k][j]))-divide[tmp];

                if(fnd>=num[tmp]){fl=false;break;}

                if(divide[tmp][fnd]!=divide[k][j]){fl=false;break;}

                if(len[tmp][fnd]*m<len[k][j]){fl=false;break;}

        }

        if(fl)ans++;

    }

    printf("%d\n",ans);

    return 0;

}