HDU——1405The Last Practice（试手map）

The Last Practice

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

Total Submission(s): 9185 Accepted Submission(s): 1947



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

60
12
-1

[align=left]Sample Output[/align]

Case 1.
2 2 3 1 5 1

Case 2.
2 2 3 1
Hint
60=2^2*3^1*5^1

本地编译器优化的太多也不好，一个语句分号前面多了个逗号竟然也可以编译，第一次妥妥的CE了

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<set>
using namespace std;
int main(void)
{
int n,i,t=0;
while (cin>>n&&n>=0)
{
t++;
if(t!=1)
cout<<endl;
map<int,int> list;
for (i=2; i<=n; i++)
{
while (n%i==0)
{
list[i]++;
n/=i;
}
}
map<int,int>::iterator it;
printf("Case %d.\n",t);
for (it=list.begin(); it!=list.end(); it++)
{
cout<<it->first<<" "<<it->second<<" ";//这题有毒竟然不按套路出牌，最后也要有空格
}
cout<<endl;
}
return 0;
}