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HDU 1405 The Last Practice

时间:2016-02-28 22:52:08      阅读:340      评论:0      收藏:0      [点我收藏+]

The Last Practice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9226    Accepted Submission(s): 1960


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
 

 

Author
lcy
 

 

Source
 
 
 
解析:给定一个正整数,将它分解为素数的乘积。
 
 
 
 1 #include <cstdio>
 2 
 3 int main()
 4 {
 5     int n,cn = 0;
 6     while(scanf("%d",&n), n>0){
 7         if(cn != 0)
 8             printf("\n");
 9         printf("Case %d.\n",++cn);
10         for(int i = 2; i*i <= n; ++i){
11             int cnt = 0;
12             while(n%i == 0){
13                 ++cnt;
14                 n /= i;
15             }
16             if(cnt != 0)
17                 printf("%d %d ",i,cnt);
18         }
19         if(n>1)
20             printf("%d 1 ",n);
21         printf("\n");
22     }
23     return 0;
24 }

 

HDU 1405 The Last Practice

原文:http://www.cnblogs.com/inmoonlight/p/5225649.html

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