hdoj Reverse Number 1266 （字符串）水

最新推荐文章于 2023-06-08 01:52:10 发布

没有能与不能只有想与不想

最新推荐文章于 2023-06-08 01:52:10 发布

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本文介绍了一种反转整数的算法实现，通过示例代码详细解释了如何处理正负整数及其末尾的零，确保反转操作正确无误。适用于初学者及需要解决类似问题的程序员。

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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

Sample Output

21
-21
2100
#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;
}