POJ 2359 Questions

Questions

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1096		Accepted: 394

Description

Holding a collegiate programming contest is a very exhausting work. There is a well-known proverb that one fool can ask so many questions that a hundred clever men will not answer. And during a collegiate programming contest questions are asked by one hundred clever people.

The jury of the Third Urals Collegiate Programming Contest being clever enough has found a simple way to make its work easier. We have invented a simple algorithm that will help us answer ALL your numerous questions! Moreover, this algorithm guarantees that the same questions will have the same answers (this would be hardly possible, if we would undertook such a task ourselves). According to this algorithm a member of the jury starts to delete characters of the question in the following order:

Starting from the first character he or she counts out N-1 characters (spaces, punctuation marks etc. are considered to be characters too) and deletes the Nth character.
If a string ends the count continues from the beginning of the string.
After deleting a character the count restarts from the character that would be the (N+1)-st in the previous count.
If the last remaining character is a question-mark ("?") then the answer to the question is "Yes". If it is a space then the answer is "No". Any other character will lead to "No comments" answer.

You should help the jury and write a program that will do a hard work of answering your questions tomorrow. The number N is secret and will not be announced even after the end of the contest. Your program should use N=1999.

For example, taking a string "Is it a good question?" (its length is 22) the characters will be counted in the following way: "Is it a good question?Is it ... quest" and "i" will be deleted. Then the count restarts from "on?Is it..." etc., until "s" will be left (thus the answer is "No comments", as usual).

Input

The input is a question, that is any text file containing at least one character (end of line is not a character). Each character of the input (excepting the ends of lines) is a part of the question.
The size of the input file is not more than 30000.

Output

The answer.

Sample Input

Sample input #1
Does the jury of this programming contest use the
algorithm described in this problem to answer my questions?

Sample input #2
At least, will anybody READ my question?

Sample input #3
This is
UNFAIR!

Sample Output

Sample output #1
Yes

Sample output #2
No

Sample output #3
No comments

 1 /* 
 2    功能Function Description:     POJ-2359
 3    开发环境Environment:          DEV C++ 4.9.9.1
 4    技术特点Technique:
 5    版本Version:
 6    作者Author:                   可笑痴狂
 7    日期Date:                      20120801
 8    备注Notes:
 9    注意：
10         本题就一组测试数据，是遇到ctrl+z后按回车结束的。
11         下边是三个AC代码
12 */
13 /*
14 代码一：
15 #include<iostream>  
16 #include<string>  
17 using namespace std;  
18 #define N 1999    
19 int main()  
20 {   
21     string s;   
22     char ch[30003];   
23     int len,i,cnt=0;     
24     while(gets(ch))    
25         s+=ch;      //string的连接，与字符数组的strcat作用相同，也是将原来的string结束符去掉再连接上后边的字符串
26     len=s.size();   
27     for(i=2;i<=len;i++)      //约瑟夫环数学解法
28         cnt=(cnt+N)%i;     
29     if(s[cnt]=='?') 
30         printf("Yes\n");    
31     else if(s[cnt]==' ') 
32         printf("No\n");   
33     else 
34         printf("No comments\n");   
35     return 0;  
36 }
37 
38 
39 /*
40 代码二：
41 #include <cstdio>
42 #include <cstring>
43 #include <iostream>
44 #include <string>
45 using namespace std;
46 string s;
47 int N = 1999;
48 char c;
49 int main() 
50 {
51     while((c = getchar()) != EOF) 
52     {
53         if(c == '\n') continue;
54         s += c;
55     }
56     int len =s.length();
57     int cnt = 0;
58     for(int i=2;i<=len;i++)
59         cnt=(cnt+N)%i;
60     if(s[cnt] == '?') 
61         printf("Yes\n");
62     else if(s[cnt] == ' ') 
63         printf("No\n");
64     else 
65         printf("No comments\n");
66     return 0;
67 }
68 */
69 
70 
71 //代码三：
72 #include<iostream>  
73 #include<string>  
74 using namespace std;  
75 #define N 1999    
76 int main()  
77 {   
78     char ch[30003],s[30003];
79     int len,i,cnt=0; 
80     s[0]='\0';
81     while(gets(ch))    
82         strcat(s,ch);      //连接
83     len=strlen(s);   
84     for(i=2;i<=len;i++)      //约瑟夫环数学解法
85         cnt=(cnt+N)%i;     
86     if(s[cnt]=='?') 
87         printf("Yes\n");    
88     else if(s[cnt]==' ') 
89         printf("No\n");   
90     else 
91         printf("No comments\n");   
92     return 0;  
93 }

功不成，身已退