题目连接：
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1151
题目描述：
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
/*问题 将一段文字的每个单词反转，但次序不变
解题思路 模拟，具体算法见注释处的坑点*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cctype>
#include<string>
#include<algorithm>
using namespace std; int main()
{
int T,t;
char str[],str1[];
string s;
scanf("%d",&T);
while(T--){
scanf("%d",&t);
getchar();
while(t--){
cin.getline(str,); int len=strlen(str);
int i;
for(i=len-;i>=;i--)
if(isalpha(str[i])) break;
str[i+]=' ';
str[i+]='\0'; char *p=NULL;
p=str; while(*p == ' ')
p++; len=strlen(p);
for(i=;i<len;i++){
if(p[i]!=' ')
{
s += p[i];
}
else
{
reverse(s.begin(),s.end());
cout<<s;
if(i != len-) printf(" ");//坑点1
s="";
}
}
cout<<endl;
s="";
}
if(T != ) printf("\n");//坑点2
}
return ;
}