【hdu - 1266 Reverse Number】

Reverse Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2924    Accepted Submission(s): 1348

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

 

 

 

 

 

// Project name : 1266
// File name    : main.cpp
// Author       : Izumu
// Date & Time  : Sun Jul  8 17:12:48 2012


#include <iostream>
#include <string>
using namespace std;

int main()
{
    int n;
    cin >> n;
    while (n--)
    {
        char s[1000];
        cin >> s;

        if (s[0] == '-')
        {
            cout << "-";
        }

        int top = 0;
        while (s[top] != '\0')
        {
            top++;
        }
        top--;

        int current = top;
        while (s[current] == '0' && current >= 0)
        {
            current--;
        }

        while (s[current] >= '0' && s[current] <= '9')
        {
            cout << s[current];
            current--;
        }

        while (s[top] == '0' && top >= 0)
        {
            cout << "0";
            top--;
        }

        cout << endl;

    }
    return 0;
}

// end 
// ism