数学概念——J - 数论，质因数分解

J - 数论，质因数分解

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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Description

Tomorrow is contest day, Are you all ready? 
We have been training for 45 days, and all guys must be tired.But , you are so lucky comparing with many excellent boys who have no chance to attend the Province-Final.

Now, your task is relaxing yourself and making the last practice. I guess that at least there are 2 problems which are easier than this problem. 
what does this problem describe? 
Give you a positive integer, please split it to some prime numbers, and you can got it through sample input and sample output. 

 

Input

Input file contains multiple test case, each case consists of a positive integer n(1<n<65536), one per line. a negative terminates the input, and it should not to be processed.

 

Output

For each test case you should output its factor as sample output (prime factor must come forth ascending ), there is a blank line between outputs.

 

Sample Input

60

12

-1

 

Sample Output

Case 1.
2 2 3 1 5 1

Case 2.
2 2 3 1

Hint

 60=2^2*3^1*5^1 

解题思路：
这个题目很简单，但是要注意一个陷井，就是输入一个数的时候，它的结束条件是负数，一定要注意
程序代码：

#include <cstdio>
#include <cstring>
using namespace std;
int b[];
int n;
void p( )
{
    int m=n;
    while()
    {
        for(int i=;i<=n;i++ )
        if(m%i==)
        {
            b[i]++;
            m/=i;
            break;
        }
        if(m==)
            break;
    }

}
int main()
{
    int Case=;
    while(scanf("%d",&n)==&&n>)
    {
        if(Case>) printf("\n");

        memset(b,,sizeof(b));
        p();
        printf("Case %d.\n",++Case);
        for(int i=;i<=n;i++)
        if(b[i]!=)
            printf("%d %d ",i,b[i]);
        printf("\n");
    }
}