1410232221-hd-Eddy's research I

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本文介绍了一个简单的素数分解算法，该算法能够将任意整数分解为素数因子的乘积形式，并提供了完整的C语言实现代码。

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Eddy's research I

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6637 Accepted Submission(s): 3975

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

题目大意

任何一个数都可以由若干个素数相等得到。现在给你一个数，求素数相乘这个式子。

解题思路

打印出素数表，然后从最小的开始判断，如果2是其一个质因数，那么下一次从i--开始，也就是还从2开始，注意这一点，因为会出现2*2*...这一情况。最后输出的时候判断一下怎么输出*就好。

代码

#include<stdio.h>
int f[70000],s[70000],num[70000];
int main()
{
	int n;
	int i,j,k;
	f[0]=f[1]=1;
	for(i=0;i<70000;i++)
	{
		if(f[i])
		    continue;
		for(j=i+i;j<70000;j+=i)
		    f[j]=1;
	}//打印数组表 
	for(i=0,j=0;i<70000;i++)
	    if(!f[i])
	    {
			s[j]=i;
			j++;
		}
	while(scanf("%d",&n)!=EOF)
	{
		j=0;
		for(i=0;s[i]<=n;i++)
		{
			if(n%s[i]==0)
			{
				num[j]=s[i];
				n/=s[i];
				j++;
				i--;//下一次也可能是这个数 
			}
		}
		for(i=0;i<j;i++)
		{
			printf("%d",num[i]);
			if(i!=j-1)//不是等于j 
			    printf("*");
		}
		printf("\n");
	}
	return 0;
}