hdoj Reverse Number 1266 （字符串）水

Reverse Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7495    Accepted Submission(s): 3234

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
#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;char s[15];int main(){int t,i,j;scanf("%d",&t);while(t--){scanf("%s",s);int l=strlen(s);if(s[0]=='-'){for(i=l-1;i>=1;i--)if(s[i]!='0')break;printf("-");for(j=i;j>=1;j--)printf("%c",s[j]);for(j=l-1;j>i;j--)printf("%c",s[j]);printf("\n");}else{for(i=l-1;i>=0;i--)if(s[i]!='0')break;for(j=i;j>=0;j--)printf("%c",s[j]);for(j=l-1;j>i;j--)printf("%c",s[j]);printf("\n");}}return 0;}