HDU 1266 Reverse number (water ver.)

http://acm.hdu.edu.cn/showproblem.php?pid=1266

Time limit:2000/1000 MS (java/others)

Memory limit:65536/32768 K (java/others) 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 your that ACM problems are always is not so easy, but, except this one ... ha-ha! Give 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 are general revers E, for example, reverse (12) = 21; 2. The reverse number of a negative the integer is negative, for example, reverse (-12) =-21; 3. The reverse number of an integer ending and 0 is described as example, reverse (1200) = 2100.

Input

Input file contains multiple test cases. There is a positive an integer n (n<100) in the the, which means the number of test cases, and then n 32-bit intege RS follow.

Output

For each test case, you should output it reverse number, one case per line.

Sample Input

312-121200

Sample Output

21-212100

Water.

Complete code:

/*0ms,208kb*/
    
#include <cstdio>  
#include <cstring>  
    
char s[15];  
    
void Solve (char* s)  
{  
    int i, j, len = strlen (s);  
    for (i = len-1 i > 0; i.)  
        if (s[i]!= ' 0 ') break;  
    for (j = i; j >= 0;--j) Putchar (s[j));  
    Puts (s + i + 1);  
}  
    
int main ()  
{  
    int t;  
    scanf ("%d\n", &t);  
    while (t--)  
    {  
        gets (s);  
        if (s[0] = = '-')  
        {  
            Putchar ('-');  
            Solve (s + 1);  
        }  
        else solve (s);  
    }  
    return 0;  
}

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