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

最新推荐文章于 2019-03-30 17:50:16 发布

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本文介绍了一个KMP算法的实际应用案例，通过判断字符串是否包含特定模式串来解决实际问题。文章详细解释了KMP算法的工作原理，并提供了一段完整的C++实现代码。

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

题意：给出五个模式串，若主串中总共只出现过一次模式串则输出YES，否则输出NO

KMP裸题

#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++; ///继续匹配 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");
    }
}