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
1 /*问题 将一段文字的每个单词反转，但次序不变 2 解题思路 模拟，具体算法见注释处的坑点*/ 3 #include<cstdio> 4 #include<iostream> 5 #include<cstring> 6 #include<cctype> 7 #include<string> 8 #include<algorithm> 9 using namespace std; 10 11 int main() 12 { 13 int T,t; 14 char str[101],str1[101]; 15 string s; 16 scanf("%d",&T); 17 while(T--){ 18 scanf("%d",&t); 19 getchar(); 20 while(t--){ 21 cin.getline(str,101); 22 23 int len=strlen(str); 24 int i; 25 for(i=len-1;i>=0;i--) 26 if(isalpha(str[i])) break; 27 str[i+1]=‘ ‘; 28 str[i+2]=‘\0‘; 29 30 char *p=NULL; 31 p=str; 32 33 while(*p == ‘ ‘) 34 p++; 35 36 len=strlen(p); 37 for(i=0;i<len;i++){ 38 if(p[i]!=‘ ‘) 39 { 40 s += p[i]; 41 } 42 else 43 { 44 reverse(s.begin(),s.end()); 45 cout<<s; 46 if(i != len-1) printf(" ");//坑点1 47 s=""; 48 } 49 } 50 cout<<endl; 51 s=""; 52 } 53 if(T != 0) printf("\n");//坑点2 54 } 55 return 0; 56 }
