hdoj 1266 Reverse Number

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

Reverse Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5182    Accepted Submission(s): 2412

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
  

#include<stdio.h>
#include<string.h>
int main()
{
	char s[100];
	int T,n,m,i,j,k;
	scanf("%d",&T);
	getchar();//注意要在这里加吸收换行符，避免gets()吸收换行 
	while(T--)
	{
		memset(s,'\0',sizeof(s));
		gets(s);
		n=strlen(s);
		for(i=n-1,m=0;i>0;i--)
		if(s[i]=='0') m+=1;//数字后边有m个零 
		else  break;//跳出后，s[i]是不是零的最高位 
		
		if(s[0]=='-')
		{//若为负数时 
		   printf("-");//先输出负号 
		   for(j=i;j>0;j--)//再从i往前输出s[],输到s[1]位 
		   printf("%c",s[j]);
		   for(j=0;j<m;j++)//最后输出m个0 
		   printf("0");
	    }
	    else
	    {//如果为正数时 
	    	for(j=i;j>=0;j--)//再从i往前输出s[],输到s[0]位
	    	printf("%c",s[j]);
	    	for(j=0;j<m;j++)//最后输出m个零 
	    	printf("0");
	    }
	    printf("\n");
	}
	return 0;
}