Reverse Number

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

Problem Description

Welcome to 2006‘4 computer college programming contest!

Specially, I give my best regards to all freshmen! You are the future of HDU ACM! And now, I must tell you that ACM problems are always not so easy, but, except this one... Ha-Ha!

Give you an integer; your task is to output its reverse number. Here, reverse number is defined as follows:

1. The reverse number of a positive integer ending without 0 is general reverse, for example, reverse (12) = 21;

2. The reverse number of a negative integer is negative, for example, reverse (-12) = -21;

3. The reverse number of an integer ending with 0 is described as example, reverse (1200) = 2100.

Input

Input file contains multiple test cases. There is a positive integer n (n<100) in the first line, which means the number of test cases, and then n 32-bit integers follow.

Output

For each test case, you should output its reverse number, one case per line.

Sample Input

3
12
-12
1200

Sample Output

21
-21
2100

Author

lcy

Source

HDU 2006-4 Programming Contest

感觉用string做比较迅速方便啦…………~~~，这题就该一次提交ac之；

AC code：

#include<iostream>
#include<string>
using namespace std;
int main()
{
	int T,n;string str;
	while(cin>>T)
	{

		for(int i=0;i<T;i++)
		{
			cin>>str;
			int len=str.size();int k=len-1;
			if(str[0]!='-')
			{
				for(int i=len-1;i>=0;i--)
					if(str[i]!='0')
					{
						k=i;
						break;
					}
					for(int j=k;j>=0;j--)
						cout<<str[j];
					for(int p=0;p<len-1-k;p++)
						cout<<0;
					cout<<endl;
			}
			else
			{
				str.erase(0,1);
				cout<<'-';
			for(int i=len-1-1;i>=0;i--)
					if(str[i]!='0')
					{
						k=i;
						break;
					}
					for(int j=k;j>=0;j--)
						cout<<str[j];
					for(int p=0;p<len-2-k;p++)
						cout<<0;
					cout<<endl;
			}
		}

	}
return 0;
}