HDU 1405  The Last Practice

The Last Practice

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit:65536/32768 K (Java/Others)

Total Submission(s): 3835   Accepted Submission(s):749

Problem Description

Tomorrow is contest day, Are you all ready?

We have been training for 45 days, and all guys must betired.But , you are so lucky comparing with many excellent boys whohave no chance to attend the Province-Final.

Now, your task is relaxing yourself and making the lastpractice. I guess that at least there are 2 problems which areeasier than this problem.

what does this problem describe?

Give you a positive integer, please split it to some primenumbers, and you can got it through sample input and sampleoutput.

 

Input

Input file contains multiple test case, each case consists ofa positive integer n(1<n<65536), oneper line. a negative terminates the input, and it should not to beprocessed.

 

Output

For each test case you should output its factor as sampleoutput (prime factor must come forth ascending ), there is a blankline 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

 

Author

lcy

 

Source

杭电ACM集训队训练赛（IV）

 

Recommend

Ignatius.L 

简单的题，但是却陷在PE不能自拔了，唉，崩溃啊，林子大了，啥鸟都有

注意两点，是输出之间加一个空行，最后的那个数据不能加空行，还有，每组数据最后是有个空格的，不要忘记加上了。

代码：

#include<stdio.h>

int prm[50000],a[70000]={0};

int main()

{

int i,j,k,n,flag,num;

k=0;

for(i=2;i<=65536;i++)

if(!a[i]){

prm[k++]=i;

for(j=i+i;j<=65536;j+=i)

a[j]=1;

}

flag=0;

while(1)

{ 

scanf("%d",&n);

if(n<0)break;

if(flag)

printf("\n");

flag++;

printf("Case %d.\n",flag);

for(i=0;i<k&&prm[i]<=n;i++)

{

num=0;

while(n%prm[i]==0){

num++;

n=n/prm[i];

}

if(num!=0){

if(n==1)printf("%d %d \n",prm[i],num);

else printf("%d %d ",prm[i],num);

}

}

}

return 0;

}