HDOJ1164

Problem Description

Eddy's interest is very extensive, recently he is interested in prime number. Eddy discover the all number owned can be divided into the multiply of prime number, but he can't write program, so Eddy has to ask intelligent you to help him, he asks you to write a program which can do the number to divided into the multiply of prime number factor .

 

Input

The input will contain a number 1 < x<= 65535 per line representing the number of elements of the set.

 

Output

You have to print a line in the output for each entry with the answer to the previous question.

 

Sample Input

119412

 

Sample Output

112*2*13*181
用Eratosthenes筛选法筛选素数
#include <stdio.h>
#include <math.h>
#define MAX 70000
int a[MAX+1];
int prime[MAX+1];
int k=0;
//素数筛选
void sushu()
{
    prime[k]=2;//放入第一个素数
    for(int i=3;i<=MAX;i+=2)
        a[i]=0;   //除2以外的素数都是奇数
    for(int i=3;i<=sqrt(MAX+0.5);i+=2)    //从3开始筛选，根据素数定理，故只用筛到sqrt(n+0.5)即可
        if(a[i]==0)  
            for(int j=i*i;j<=MAX;j+=i*2)  //除2之外的偶数均以筛过，故可直接从i*i开始筛，j每次循环加2*i，保持j是奇数
                a[j]=1;
    for(int i=3;i<=MAX;i+=2)
        if(a[i]==0)
            prime[++k]=i;
}
int main()
{
    shaixuan();
    int x;
    while(scanf("%d",&x)!=-1)
    {
        int i=0;
        for(int i=0;i<k;i++)
        {
            //如果是素数，直接输出，换行结束
            if(x==prime[i])
            {
                printf("%d",x);
                printf("\n");
                break;
            }
            //如果不是素数，则输出“素数*”，并且自身除去素数
            if(x%prime[i]==0)
            {
                printf("%d*",prime[i]);
                x/=prime[i];
                //i需要自减，为了不再次重头再查素数表，可能当前素数有几个，比如12=2*2*3
                i--;
            }
        }
    }
    return 0;
}