SGU 404 Fortune-telling with camomile

404. Fortune-telling with camomile

Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard




Masha loves Petya. The following question gives her no rest: does Petya love her too? The best way to find this out is a fortune-telling. There are plenty ways of fortune predicting, but Masha prefers fortune-telling with camomile more than others. It's rules are simple. You should take camomile into the right hand and start picking petals one by one. After each petal you should pronounce one phrase from the predefined list. Such phrases like "loves", "doesn't love", "loves sincerely", "doubts", "wants to date", "laughs" are usually used. Phrases are pronounced from the first to the last. The list of phrases is cyclic, so after the last phrase you should pronounce the first one. The phrase that you pronounce after the last petal will be an answer.

Since Masha doesn't want to go to the forest and look for camomiles, she asks you to write the program which will simulate the process.

Input

First line of the input file contains two integer numbers N and M (1 ≤ N ≤ 100, 1 ≤ M ≤ 100), the number of petals and the number of phrases. Each of the following M lines contains one phrase. Phrases consist only of latin letters and their lengths are between 1 and 100.

Output

Output the resulting phrase.

Example(s)
sample input
sample output
6 4
loves
doesnt
sincerely
doubts
doesnt

 

sample input
sample output
9 3
loves
doesnt
hates
hates
 1 #include<iostream>  
 2 #include<string.h>  
 3 #include<stdio.h>  
 4 #include<ctype.h>  
 5 #include<algorithm>  
 6 #include<stack>  
 7 #include<queue>  
 8 #include<set>  
 9 #include<math.h>  
10 #include<vector>  
11 #include<map>  
12 #include<deque>  
13 #include<list>  
14 using namespace std;
15 char str[110][110];
16 int main()
17 {
18     int n,m;
19     while(scanf("%d%d",&n,&m)==2){
20         for(int i=1;i<=m;i++){
21             scanf("%s",str[i]);
22         }
23         int ans=n%m;
24         if(ans==0) printf("%s\n",str[m]);
25         else printf("%s\n",str[ans]);
26     }
27     return 0;
28 }
View Code

 

posted @ 2014-07-31 09:45  qscqesze  阅读(162)  评论(0编辑  收藏  举报