题干：

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

解题报告：

   别忘特盘一下n==0的情况，因为这种情况fit函数处理不了。

AC代码：

#include<cstdio>
#include<queue>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAX = 100000 +5;
int a[1000];
int n,x,p;
void fit(int x) {
	while(x) {
		a[++p] = x%10;
		x/=10;
	}
}
int main()
{
	int t,flag;
	cin>>t;
	while(t--) {
		p = 0;
		flag = 0;
		scanf("%d",&n);
		if(n<0) flag=1,n=-n;
		if(n == 0) {
			puts("0");continue;
		}
		fit(n);
		int i=0;
		for(i = 1; i<=p; i++) {
			if(a[i]!=0) break;
		}
		int cnt0 = i-1;
		if(flag == 1) putchar('-');
		for(int j = i; j<=p; j++) {
			printf("%d",a[j]);
		}
		for(int j = cnt0; j>0; j--) putchar('0');
		printf("\n");
		
	}
	return 0;
}