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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>

int main(){
    //char array[100];
    //for(int i =0; i<=99; i++){
    //    array[i]=a;
    //}
    int a=0;
    int b=0;
    int c=0;
    int d=0;
    int e=0;
    int f=0;
    int g=0;
    int h=0;
    int suiso=0;
    char array[101];
    int j=0;
    int mojiretu;
    scanf("%s",array);
    for(j=0; array[j]!='\0';j++);
    //printf("%d\n",j);
    mojiretu=j; 
    for(j=0;j<=99;j++){
        //array[j]='a';
        //printf("%s",array[j]);
    }    
    for(j=0;j<=mojiretu-1;j++){
        if(array[j]=='k' && a==0){
            a++;
            suiso=1;
        }
        else if(array[j]=='e' && b==0 && a==1){
            b++;
            suiso=1;
        }
    
        else if(array[j]=='y' && c==0 && b==1){
            c++;
            suiso=1;
        }
        else if(array[j]=='e' && d==0 && c==1){
            d++;
            suiso=1;
        }
        else if(array[j]=='n' && e==0 && d==1){
            e++;
            suiso=1;
        }
        else if(array[j]=='c' && f==0 && e==1){
            f++;
            suiso=1;
        }
        else if(array[j]=='e' && g==0 && f==1){
            g++;
            suiso=1;
        }
        else{
            if(suiso==1||j==0){
                h++;
            }
            suiso=0;
        }
    }
    //printf("\n%d %d %d %d %d %d %d %d %d\n",a,b,c,d,e,f,g,h,j);
    if(a>0 && b>0 && c>0 && d>0 && e>0 && f>0 && g>0 && h<2){
        printf("YES\n");
    }
    else{
        printf("NO\n");
    }
    return 0;
}