hdu1266 Reverse Number

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

312-121200

 

Sample Output

21-212100

 

Author

lcy

 

Source

HDU 2006-4 Programming Contest

C语言AC代码

#include<stdio.h>#include<string.h>#define Max 666666char a[Max];int main(){    int i,j,k,l,n;    scanf("%d",&n);    while(n--)    {        scanf("%s",&a);        l=strlen(a);        for(k=l-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<l;i++)            {                printf("0");            }            printf("\n");        }        else        {            for(i=k;i>=0;i--)            {                printf("%c",a[i]);            }            for(i=k+1;i<l;i++)            {                printf("0");            }            printf("\n");        }    }    return 0;}

一望而知。