Reverse Number

Reverse Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3872    Accepted Submission(s): 1784

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

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std ;
int main(void)
{
    int T;
    cin >> T ;
    while(T--){
        int n;
        cin >> n ;
        char str[102];
        sprintf(str,"%d",n);
        int flag = 0 ;
        if(n<0)
            cout<<"-" ;
        reverse(str,str+strlen(str));

        int cnt = 0 , i , j ;
        for( i = 0 ; i < strlen(str) ; i++)
            if(str[i]=='0')
                cnt ++ ;
            else
                break;
        for(  ; i < strlen(str) ; i ++)
            if(str[i]>='0'&&str[i]<='9')
                cout<<str[i];
        for( j = 0 ; j < cnt ; j ++)
            cout<<"0";
        cout<<endl;
    }
    return 0;
}