HDU 4730 We Love MOE Girls

痛定思痛。 2022-09-28 11:46 12阅读 0赞

# We Love MOE Girls #

**Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 3312    Accepted Submission(s): 1871**

Problem Description

Chikami Nanako is a girl living in many different parallel worlds. In this problem we talk about one of them.  
In this world, Nanako has a special habit. When talking with others, she always ends each sentence with "nanodesu".  
There are two situations:  
If a sentence ends with "desu", she changes "desu" into "nanodesu", e.g. for "iloveyoudesu", she will say "iloveyounanodesu". Otherwise, she just add "nanodesu" to the end of the original sentence.  
Given an original sentence, what will it sound like aften spoken by Nanako?

Input

The first line has a number T (T <= 1000) , indicating the number of test cases.  
For each test case, the only line contains a string s, which is the original sentence.  
The length of sentence s will not exceed 100, and the sentence contains lowercase letters from a to z only.

Output

For every case, you should output "Case \#t: " at first, without quotes. The  *t* is the case number starting from 1. Then output which Nanako will say.

Sample Input

2
    ohayougozaimasu
    daijyoubudesu

Sample Output

Case #1: ohayougozaimasunanodesu
    Case #2: daijyoubunanodesu

Source

[2013 ACM/ICPC Asia Regional Chengdu Online][2013 ACM_ICPC Asia Regional Chengdu Online]

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    #include <string>
    using namespace std;
    
    int main()
    {
        int T;
        cin>>T;
        int tot = 0;
        while(T--)
        {
            string a;
            cin>>a;
            cout<<"Case #"<<++tot<<": ";
            int len = a.length();
             if(a[len-1] == 'u' && a[len-2] == 's' && a[len-3] == 'e' && a[len-4] == 'd' &&
               a[len-5] == 'o' && a[len-6] == 'n' && a[len-7] == 'a' && a[len-8] == 'n')
                {
                    cout<<a<<endl;
                }
                else if(a[len-1] == 'u' && a[len-2] == 's' && a[len-3] == 'e' && a[len-4] == 'd')
                {
                    for(int i=0;i<len -4;i++)
                    {
                        cout<<a[i];
                    }
                    cout<<"nanodesu"<<endl;
                }
                else
                {
                    cout<<a<<"nanodesu"<<endl;
                }
    
    }
      return 0;
    }

[2013 ACM_ICPC Asia Regional Chengdu Online]: http://acm.hdu.edu.cn/search.php?field=problem&key=2013+ACM%2FICPC+Asia+Regional+Chengdu+Online&source=1&searchmode=source