HDU 1266 Reverse Number （很简单的水题）

Reverse Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2047    Accepted Submission(s): 926

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

 

/*
HDU 1266
水题，但是要用字符串去存储，并不像题目说的是整型的
数据很大 
*/
#include<stdio.h>
#include<string.h>
int main()
{
    char m[100];
    long long x;
    int T,i;
    int t;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",&m);
        int k=strlen(m);         
        for(t=k-1;t>=0;t--)
           if(m[t]!='0')  break;
        if(m[0]=='-')
        {
            printf("-");
            for(i=t;i>=1;i--)
              printf("%c",m[i]);
            for(i=t+1;i<k;i++)
              printf("0");
        }    
        else
        {

            for(i=t;i>=0;i--)
              printf("%c",m[i]);
            for(i=t+1;i<k;i++)
              printf("0");
        }    
        printf("\n");
    }  
    return 0;  
}