hdoj 1405 The Last Practice

The Last Practice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9037    Accepted Submission(s): 1905

Problem 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

 

题意：输入一个n，分解出其质因子，并输出其中有多少个相同的质因子

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<math.h>
#define MAX 1100
using namespace std;
int a[MAX],b[MAX];
int main()
{
	int k,n,m,j,i;
	k=1;
	while(scanf("%d",&n)&&n>=0)
	{
		if(k>1)
		printf("\n");
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		printf("Case %d.\n",k++);
		m=n;j=0;
		for(i=2;i<=n;i++)
		{
			int sum=0;
			if(m%i==0)
			{
				while(m%i==0)
				{
					m/=i;
					sum++;
				}
				a[j]=i;
				b[j++]=sum;
			}
			if(m==1)
			    break;
		}
		for(i=0;i<j;i++)
		    printf("%d %d ",a[i],b[i]);
		printf("\n");
	}
	return 0;
}