HDU1266-Reverse Number

Reverse Number

                                                                           Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                      Total Submission(s): 9351    Accepted Submission(s): 4100

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  

Recommend

lxj

 

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>

using namespace std;

int main()
{
    char ch[50];
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",ch);
        int len=strlen(ch);
        if(ch[0]=='-') printf("%c",ch[0]);
        int i=len-1;
        while(ch[i]=='0') i--;
        int j=ch[0]=='-'?1:0;
        for(int k=i;k>=j;k--) printf("%c",ch[k]);
        i++;
        while(i<len) printf("0"),i++;
        printf("\n");
    }
    return 0;
}