CF#877 A. Alex and broken contest（KMP）

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

Alex_and_broken_contest

output

NO

input

NikitaAndString

output

YES

input

Danil_and_Olya

output

NO

題意：給出五個模式串，若主串中總共只出現過一次模式串則輸出YES，否則輸出NO

KMP裸題

#include <bits/stdc++.h>
using namespace std;

const int N = 1000 + 10;
int t, n, m;
char s[N], c[10][N];
int nex[N];

void get_nex(char b[], int lb){
    int i = 0, j = -1;
    nex[0] = -1;
    while(i < lb){
        if(j == -1 || b[i] == b[j]){
            i++, j++;
            nex[i] = j;
        }
        else j = nex[j];
    }
}

int kmp(char a[], char b[]){
    int la = strlen(a);
    int lb = strlen(b);
    int sum = 0;
    get_nex(b, lb);
    int i = 0, j = 0;
    while(i < la){
        if(j == -1 || a[i] == b[j]){
            i++, j++;
        }
        else j = nex[j];
        if(j == lb) sum++; ///繼續匹配 j = 0;
    }
    return sum;
}

int main(){
    strcpy(c[0], "Danil");
    strcpy(c[1], "Olya");
    strcpy(c[2], "Slava");
    strcpy(c[3], "Ann");
    strcpy(c[4], "Nikita");
    while(scanf("%s", s) == 1){
        int sum = 0;
        for(int i = 0; i < 5; i++){
            sum += kmp(s, c[i]);
        }
        printf(sum == 1 ? "YES\n" : "NO\n");
    }
}