HDU 1405 The Last Practice(暴力枚举)

The Last Practice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9522    Accepted Submission(s): 2034

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

​​杭电ACM集训队训练赛（IV） ​​

题解：暴力枚举2到sqrt（n）,记录次数，输出，注意输入输出格式.....（这个坑）

AC代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<algorithm>
#include<time.h>
typedef long long LL;
using namespace std;
int main()
{
    int n;
    int m=1;
    while(cin>>n)
    {
        int j;
        if(n<0)break;
        if(m>1)cout<<endl;
        cout<<"Case "<<m++<<"."<<endl;
        if(n==1)cout<<"1 1 ";
        for(int i=2;i<=sqrt(n);i++)
        {
            j=0;
            while(n%i==0)
            {
                n=n/i;
                j++;
            }
            if(j)
            cout<<i<<" "<<j<<" ";
        }
        if(n!=1)
        cout<<n<<" "<<"1"<<" ";
        cout<<endl;
    }
    return 0;
}