Codeforces Round #442 (Div. 2)　A. Alex and broken contest

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本文解析了一个CodeForces上的简单字符串匹配题目，介绍了如何通过C++及C语言实现精确匹配特定字符串的功能，并提供了完整的代码示例。

http://codeforces.com/problemset/problem/877/A

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

input

NikitaAndString

output

YES

input

Danil_and_Olya

output

题意

给出了五个朋友的名字，当且仅当其名称中包含Alex的一个朋友的名字一次时，输出ＹＥＳ否则输出ＮＯ

题解

很简单的题，不会ｃ＋＋的话有些麻烦

c++CODE

#include "bits/stdc++.h"
using namespace std;
string a;
int i,cnt;
vector <string> v={"Danil","Olya","Slava","Ann","Nikita"};

int main()
{
    cin>>a;

    for(i=0;i<a.size();i++)
        for(auto j:v)
            cnt+=(a.substr(i,j.size())==j);

    if(cnt==1) cout<<"YES";
    else cout<<"NO";
}

C CODE

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
int main()
{
    char s1[10]="Danil",s2[10]="Olya",s3[10]="Slava",s4[10]="Ann",s5[10]="Nikita";
    char ch[110];
    scanf("%s",ch);
    int l=strlen(ch);
    int sum=0;
    for(int i=0;i<l;i++)
    {
        int flag=0;
        if(ch[i]=='D')
        {
            for(int j=i;j<=i+5;j++)
                if(ch[j]!=s1[j-i])
                    flag=1;
            if(flag)
                continue;
            else
                sum++;
        }
        if(ch[i]=='O')
        {
            for(int j=i;j<=i+4;j++)
                if(ch[j]!=s2[j-i])
                    flag=1;
            if(flag)
                continue;
            else
                sum++;
        }
        if(ch[i]=='S')
        {
            for(int j=i;j<=i+5;j++)
                if(ch[j]!=s3[j-i])
                    flag=1;
            if(flag)
                continue;
            else
                sum++;
        }
        if(ch[i]=='A')
        {
            for(int j=i;j<=i+3;j++)
                if(ch[j]!=s4[j-i])
                    flag=1;
            if(flag)
                continue;
            else
                sum++;
        }
        if(ch[i]=='N')
        {
            for(int j=i;j<=i+6;j++)
                if(ch[j]!=s5[j-i])
                    flag=1;
            if(flag)
                continue;
            else
                sum++;
        }
    }
    if(sum==1)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
}