ZOJ 1151 - Word Reversal

最新推荐文章于 2019-01-30 21:15:23 发布

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本文详细介绍了一种名为WordReversal的算法，该算法能在保持单词顺序不变的情况下，实现单词内部字符的反转。通过分析代码，我们了解到算法的具体实现过程，包括字符串处理和字符反转的技巧。此外，还提到了在实现过程中可能遇到的一些问题，如数组初始化、空格处理等，并提供了完整的代码示例。

Word Reversal

Time Limit: 2 Seconds Memory Limit: 65536 KB

For each list of words, output a line with each word reversed without changing the order of the words.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input

You will be given a number of test cases. The first line contains a positive integer indicating the number of cases to follow. Each case is given on a line containing a list of words separated by one space, and each word contains only uppercase and lowercase letters.

Output

For each test case, print the output on one line.

Sample Input

3
I am happy today
To be or not to be
I want to win the practice contest

Sample Output

I ma yppah yadot
oT eb ro ton ot eb
I tnaw ot niw eht ecitcarp tsetnoc

忽略了每次b数组的初始问题，忽略了如果是最后一个单词后面没有空格的情况，还有把空格也存进b数组的问题。

代码：

#include<stdio.h>
#include<string.h>
char a[1010],b[1010];
int change1()
{
	int i,len=0;
	len=strlen(b);
	for(i=len-1;i>=0;i--)
		printf("%c",b[i]);
}
int change()
{
	int i,j=0;
	for(i=0;a[i]!='\0';i++)
	{
		if(a[i]==' ')
		{
			b[j]='\0';
			change1();
			j=0;
			printf(" ");
		}
		else	
			b[j++]=a[i];
		if(a[i+1]=='\0'&&a[i]!=' ')
		{
			b[j]='\0';
			change1();
		}
	}
	
	
}
int main()
{
	int t,n,i;
	char ch;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		getchar();
		while(n--)
		{
			gets(a);
			change();
			printf("\n");
		}
		if(t)
			printf("\n");
	}
	return 0;
}