杭电1266 Reverse Number

Reverse Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3771    Accepted Submission(s): 1736

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

 

如题中描述，这是一道简单题！但开始做的时候考虑的不够周全，遗漏了0出现在整数非末端的情况，导致一次WA！改进后AC，代码如下：

#include <stdio.h>
#include <stdlib.h>

int reverse( int x )
{
    int cnt  = 0;
    if( x < 0 )
    {
        printf( "-" );
        x = -x;
    }
    else if( x == 0 )
    {    
         printf( "%d\n",x );
         return 0;
    }
    while( x % 10 == 0 )
    {
           x = x / 10;
           cnt++;
    }
    while( x != 0 )
    {
           printf( "%d", x % 10 );
           x = x / 10;
    }
    while( cnt-- )
           printf( "%d", 0 );
    printf( "\n" );
    return 0;
}

int main(int argc, char *argv[])
{
    int n, a;
    scanf( "%d", &n );
    while( n-- )
    {
           scanf( "%d", &a );
           reverse(a);      
    }
  
//  system("PAUSE");    
  return 0;
}