zoj 1151 Word Reversal(字符串操作模拟)

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 ;
 }