zoj1151 I - Word Reversal 【简单字符串处理】【第三周练习】【the first day】

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
1

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

题意：将每组字符串的单词一一翻转输出，t组样例，每组样例包括n行字符串，每组样例间有空行。
思路：确定每个单词在字符数组中的起始位置和终止位置，将该区间内的单词翻转后存入数组，具体细节见注释。

总结：老师应该是看我们被上次的练习打击的太惨了，所以这次放一道签到题在这里吧，细心一点应该可是实现1A，可是我就是那个不细心的人呐，好好读题行不行，每组样例之间的空行呢！！

#include<stdio.h>
#include<string.h>
#define N 1000
struct node{
char str
;
}sum[N+10];

void Reverse(int start,int end,int number)
{
char s
;
int j = 0;
for(int i = end; i >= start; i--)
s[j++] = sum[number].str[i];
s[j] = '\0';//将从start开始到end结束的字符段翻转
for(int i = start,k= 0; i <= end&&s[k]!='\0'; i ++,k++)
sum[number].str[i] = s[k];//翻转后存入原字符组
return;
}

int main()
{
int t,l,n,start,end,flag=0;
scanf("%d",&t);
while(t--)
{
if(flag)
printf("\n");//认真读题！！
flag = 1;
scanf("%d",&n);
getchar();
for(int i = 0; i < n; i ++)
gets(sum[i].str);
for(int i = 0; i < n; i ++)
{
start = end=0;
while(sum[i].str[start]!='\0')
{
if(sum[i].str[end]==' '||sum[i].str[end] =='\0')//遇到空格或者终止符说明为一个单词
{
Reverse(start,end-1,i);//翻转函数
if(end == '\0')
break;//字符串结尾直接跳出循环
start = end+1;//跳过空格开始翻转下一个字符段
}
end ++;
}
}
for(int i = 0; i < n; i ++)
puts(sum[i].str);
}
return 0;
}