Hduoj1266【水题】

/*Reverse Number
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5182    Accepted Submission(s): 2412


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 
*/
#include<stdio.h>
#include<string.h>
int main()
{
	char s[25],ch;
	__int64 i, j, k, m ,n, sum;
	scanf("%I64d",&n);
	getchar();
	while(n--)
	{
		
		sum = 0;
		gets(s);
		m = strlen(s);
		if( strchr(s, '-') )
		{
			k = 0;
			j = 0;
			for(i = 1; i < m; i++)
			{
				if(s[i] != '0' && !k )
				k = i;
				if(s[m-i] != '0' && !j )
				j = m-i;
				if( k && j)
				break;
			}
			for(i = k; i <= (j+k)/2; i++)
			{
				ch = s[i];
				s[i] = s[ j+k - i];
				s[j+k-i] = ch;
			}
			if(!k)
			k = 1;
			for(i = k; i < m; i++)
			{
				sum = sum * 10 + (s[i] - 48);
			}
			if(sum == 0)
			printf("0\n");
			else
			printf("-%I64d\n", sum);
		}
		else
		{
			k = -1;
			j = -1;
			for(i = 0; i < m; i++)
			{
				if(s[i] != '0' && k == -1 )
				k = i;
				if(s[m-1-i] != '0' && j == -1)
				j = m-1-i;
				if( k != -1 && j != -1)
				break;
			}
			for(i = k; i <= (j+k)/2 ; i++)
			{
				ch = s[i];
				s[i] = s[ j+k - i];
				s[j+k-i] = ch;
			}
			if(k == -1)
			k = 0;
			for(i = k; i < m; i++)
			{
				sum = sum * 10 + (s[i] - 48);
			}
			printf("%I64d\n", sum);
		}
	}
	return 0;
}