Codeforces 877A - Alex and broken contest

A. Alex and broken contest

time limit per test

2 seconds

memory limit per test

256 megabytes

 

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

 思路：第一种直接暴力匹配，第二种利用c++的substr（）函数

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    string s[5]={"Danil", "Olya", "Slava", "Ann", "Nikita"};
    string ss;
    while(getline(cin, ss))
    {
        int cnt = 0;
        for(int i = 0;i < ss.length();i++)
        {
            for(int j = 0;j < 5;j++)
                if(ss[i] == s[j][0])
                {
                    int len = s[j].length();
                    int tmp = i;
                    bool flag = true; 
                    for(int k = 0;k < len;k++)
                    {
                        if(ss[tmp] != s[j][k])
                            flag = false;
                        tmp++;
                    }
                    if(flag)
                        cnt++;
                }
            if(cnt >= 2)
            {
                printf("NO\n");
                break;
            }
        }
        if(cnt == 1)
            printf("YES\n");
        else if(cnt == 0)
            printf("NO\n");
    }
} 

第二种

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <cstring>
using namespace std;
typedef long long LL;
string str[] = { "Danil", "Olya", "Slava", "Ann", "Nikita" };
string S;
int main()
{
    cin >> S;
    int flag = 0;
    for (int i = 0; i < 5; i++)
    {
        for (int j = 0; j < S.size(); j++)
        {
            if (S.substr(j, str[i].size()) == str[i])//依次遍历字符串S,找到该串j位置到str[i]长度位置中符合str[i]的字符串
            {
                flag++;
            }
        }
    }
    if (flag == 1)
    {
        cout<<"YES"<<endl;
    }
    else
    {
        cout<<"NO"<<endl;
    }
    return 0;
}