CodeForces 877A Alex and broken contest KMP Getting Started

A. Alex and broken contest

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Alex was creating a contest about his friends, but accidentally deleted it. Fortunately, all the problems were saved, but now he needs to find them among other problems.

But there are too many problems, to do it manually. Alex asks you to write a program, which will determine if a problem is from this contest by its name.

It is known, that problem is from this contest if and only if its name contains one of Alex's friends' name exactly once. His friends' names are "Danil", "Olya", "Slava", "Ann" and "Nikita".

Names are case sensitive.

Input

The only line contains string from lowercase and uppercase letters and "_" symbols of length, not more than 100 — the name of the problem.

Output

Print "YES", if problem is from this contest, and "NO" otherwise.

Examples

input

Copy

Alex_and_broken_contest

output

Copy

NO

input

Copy

NikitaAndString

output

Copy

YES

input

Copy

Danil_and_Olya

output

Copy

NO

The key is to query if there is only one person's name appearing && this person's name only appearing once.

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;

void compute_prefix(int *next,char *p)
{
	int n,k;
	n=strlen(p);
	next[0]=next[1]=0;
	k=0;
	for(int i=2;i<=n;i++)
	{
		while(k&&p[k]!=p[i-1])
		k=next[k];
		if(p[k]==p[i-1])
		k++;
		next[i]=k;
	}
}

int kmp_match(char *text,char *p,int *next)
{
	int ans=0,j=0;
	int text_len=strlen(text);
	int p_len=strlen(p);
	for(int i=0;i<text_len;i++)
	{
		while(j&&text[i]!=p[j])
		j=next[j];
		if(text[i]==p[j])
		j++;
		if(j==p_len)
		years++;
	}
	return ans;
}






char s[200];
int next[200];
intmain()
{
	
	while(scanf("%s",s)!=EOF)
	{
		char name[5][10]={"Danil", "Olya", "Slava", "Ann" , "Nikita"};
		int flag=0;
		for(int i=0;i<5;i++)
		{
			flag+=kmp_match(s,name[i],next);
	
		}
		if(flag==1)
		printf("YES\n");
		else
		printf("NO\n");
	}
	
	return 0;
}