Problem:
Score : 200 points

Problem Statement

A string is called a KEYENCE string when it can be changed to keyence by removing its contiguous substring (possibly empty) only once.

Given a string S consisting of lowercase English letters, determine if S is a KEYENCE string.

Constraints

The length of S is between 7 and 100 (inclusive).

S consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If S is a KEYENCE string, print YES; otherwise, print NO.

Sample Input 1

keyofscience

Sample Output 1

YES

keyence is an abbreviation of key of science.

Sample Input 2

mpyszsbznf

Sample Output 2

NO

Sample Input 3

ashlfyha

Sample Output 3

NO

Sample Input 4

keyence

Sample Output 4

YES

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
int main(void){
    string s,p = "keyence";
    bool b = false;
    cin >> s;
    for(int i = 0;i < s.length();i++){
        if(s[i] == 'k' && s[s.length()-1] == 'e' && s[s.length()-2] == 'c' &&s[s.length()-3] == 'n' &&s[s.length()-4] == 'e' &&s[s.length()-5] == 'y' &&s[s.length()-6] == 'e'){
            b = true;
        }
        if(s[i] == 'k' && s[i+1] == 'e' && s[s.length()-1] == 'e' && s[s.length()-2] == 'c' &&s[s.length()-3] == 'n' &&s[s.length()-4] == 'e' &&s[s.length()-5] == 'y'){
           b = true;
        }
         if(s[i] == 'k' && s[i+1] == 'e' && s[i+2] == 'y'&& s[s.length()-1] == 'e' && s[s.length()-2] == 'c' &&s[s.length()-3] == 'n' &&s[s.length()-4] == 'e'){
            b = true;
        }
         if(s[i] == 'k' && s[i+1] == 'e' && s[i+2] == 'y'&& s[i+3] == 'e'&& s[s.length()-1] == 'e' && s[s.length()-2] == 'c' &&s[s.length()-3] == 'n'){
            b = true;
        }
         if(s[i] == 'k' && s[i+1] == 'e' && s[i+2] == 'y'&& s[i+3] == 'e'&& s[i+4] == 'n'&& s[s.length()-1] == 'e' && s[s.length()-2] == 'c'){
            b = true;
        }
         if(s[i] == 'k' && s[i+1] == 'e' && s[i+2] == 'y'&& s[i+3] == 'e'&& s[i+4] == 'n'&& s[i+4] == 'c'&& s[s.length()-1] == 'e'){
           b = true;
        }
        if(s[i] == 'k' &&s[i+1] == 'e' &&s[i+2] == 'y' &&s[i+3] == 'e' &&s[i+4] == 'n' &&s[i+5] == 'c' &&s[i+6] == 'e'){
           b = true;
        }
        if(s[s.length()-7] == 'k' &&s[s.length()-6] == 'e' &&s[s.length()-5] == 'y' &&s[s.length()-4] == 'e' &&s[s.length()-3] == 'n' &&s[s.length()-2] == 'c' &&s[s.length()-1] == 'e'){
            b = true;
        }
    }
    if(b){
        cout << "YES" << endl;
        return 0;
    }else{
        cout << "NO"<< endl;
        return 0;
    }
}