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  • Word Reversal(string)

    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 #include<stdio.h>
     2 #include<iostream>
     3 #include<algorithm>
     4 #include<string>
     5 using namespace std;
     6 
     7 int main()
     8 {
     9     int t;
    10     cin >> t;
    11     getchar();
    12     while(t--)
    13     {
    14         int cas;
    15         cin >> cas;
    16         getchar();
    17         while(cas--)
    18         {
    19             string s, t;
    20             getline(cin, s);
    21             t.clear();
    22             int  flag = 0;
    23             for(int i = 0; i < s.length(); i++)
    24             {
    25                 if(s[i] != ' ' && i <= s.length() - 1)
    26                     t += s[i];
    27                 else
    28                 {
    29                     reverse(t.begin(), t.end());
    30                     if(flag)
    31                         cout << " ";
    32                     cout << t;
    33                     flag = 1;
    34                     t.clear();
    35                 }
    36             }
    37             reverse(t.begin(), t.end());
    38             cout << " " << t << endl;
    39         }
    40         return 0;
    41     }
    42 
    43 }
    View Code
    永远渴望,大智若愚(stay hungry, stay foolish)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/8666469.html
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