杭电oj-1164-Eddy's research I

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

11

9412

Sample Output

11

2*2*13*181

题目的意思是给你一个整数x(1<x<=65535),让你写出x的素数因数乘积式，如果本身x是素数，则直接输出x。

分析：这是一道简单的素数筛选题，用Eratosthenes优化筛法筛出素数表，剩下查表就ok了，注意不要忘了有的数有多个相同素数因子。

代码如下：

#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 shaixuan()

{

    prime[k]=2;

    for(int i=3;i<=MAX;i+=2)

        a[i]=0;

    for(int i=3;i<=sqrt(MAX+0.5);i+=2)

        if(a[i]==0)

            for(int j=i*i;j<=MAX;j+=i*2)

                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9=3*3

                i--;

            }

        }

    }

    return 0;

}