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hdu 4730 We Love MOE Girls
We Love MOE Girls
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1105 Accepted Submission(s): 674
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?
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.
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. Thet 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
题意:水题。额,只要看最后几位是否满足这种情况,如果不满足的话直接加上这个字符。否则将前面几个字符给删掉,然后继续加。
#include<iostream> #include<cstdio> #include<cstring> #include<string> using namespace std; int main() { int t,i,j; string s; int step=1; scanf("%d",&t); while(t--) { cin>>s; int len=s.length(); if(len>=4&&s[len-4]=='d'&&s[len-3]=='e'&&s[len-2]=='s'&&s[len-1]=='u') { s[len-4]='n'; s[len-3]='a'; s[len-2]='n'; s[len-1]='o'; s+="desu"; } else s+="nanodesu"; cout<<"Case #"<<step<<": "<<s<<endl; step++; } return 0; }
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