hdoj-1266-Reverse Number

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

312-121200 

 

Sample Output

21-212100 

 

给你一个数，把它倒过来就好了。但是不能有前导零。

字符串去做啊，记录下有几个零和零的位置然后反着输出来就好了

#include<stdio.h>#include<string.h>int main(){char a[10000];    int t, n, i, x,k;    scanf("%d", &t);    while( t-- )    {        scanf("%s", &a);        x = strlen(a);        for(k = x - 1; k >= 0; k--)        {            if(a[k] != '0')                break;        }        if(a[0] == '-')        {            printf("-");            for(i = k; i >= 1; i--)                printf("%c", a[i]);            for(i = k + 1; i < x; i++)                printf("0");                   printf("\n");        }        else        {            for(i = k; i >= 0; i--)                printf("%c", a[i]);           for(i = k + 1; i < x; i++)                printf("0");    printf("\n");        }    }    return 0;}