Codeforces Round #442 A Alex and broken contest【字符串/常量数组/string类】

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

【题意】：给出字符串，问能否找出5个名字中的某一个恰出现一次，重复出现的名字也不行。
【分析】：常量字符数组的妙用，注意发现某名字后还要统计次数。标记满足条件者。
【代码】：

#include <bits/stdc++.h>
using namespace std;
int fun(const std::string& str, const std::string& sub)
    {
        int num = ;
        size_t len = sub.length();
        if (len == )len=;//应付空子串调用
        for (size_t i=; (i=str.find(sub,i)) != std::string::npos; num++, i+=len);
        return num;
    }
int fun(string s, string t) {//
    int res = ;
    for (int i = ; i < (int)s.size(); i++) {
        if (s.substr(i, t.size()) == t) res++;
    }
    return res;
}
int main()
{
    string s[] = {"Danil", "Olya", "Slava", "Ann", "Nikita"};
    string t;
    cin>>t;
    int cnt=;
    for(int i=;i<;i++)
    {
        cnt+=fun(t,s[i]);
    }
    if(cnt==)
        cout<<"YES";
    else
        cout<<"NO";
    return ;
}

strstr函数

#include <bits/stdc++.h>
using namespace std;
int main() {
    string s;
    string a[] = {"Danil", "Olya", "Slava", "Ann", "Nikita"};
    cin >> s;
    int cnt = ;
    for (string i: a) {
        size_t pos = s.find(i, );
        while(pos != string::npos)
        {
            cnt++;
            pos = s.find(i, pos + );
        }
    }

    printf("%s\n", cnt ==  ? "Yes" : "No");
    return ;
}

find函数