CF 538A(Cutting Banner-暴力找切割点)

A. Cutting Banner

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A large banner with word CODEFORCES was ordered for the 1000-th onsite round of Codeforcesω that takes place on the Miami beach. Unfortunately, the company that made the banner mixed up two orders and delivered somebody else's banner that contains someone else's word. The word on the banner consists only of upper-case English letters.

There is very little time to correct the mistake. All that we can manage to do is to cut out some substring from the banner, i.e. several consecutive letters. After that all the resulting parts of the banner will be glued into a single piece (if the beginning or the end of the original banner was cut out, only one part remains); it is not allowed change the relative order of parts of the banner (i.e. after a substring is cut, several first and last letters are left, it is allowed only to glue the last letters to the right of the first letters). Thus, for example, for example, you can cut a substring out from string 'TEMPLATE' and get string 'TEMPLE' (if you cut out string AT), 'PLATE' (if you cut out TEM), 'T' (if you cut out EMPLATE), etc.

Help the organizers of the round determine whether it is possible to cut out of the banner some substring in such a way that the remaining parts formed word CODEFORCES.

Input

The single line of the input contains the word written on the banner. The word only consists of upper-case English letters. The word is non-empty and its length doesn't exceed 100 characters. It is guaranteed that the word isn't word CODEFORCES.

Output

Print 'YES', if there exists a way to cut out the substring, and 'NO' otherwise (without the quotes).

Sample test(s)

input

CODEWAITFORITFORCES

output

YES

input

BOTTOMCODER

output

NO

input

DECODEFORCES

output

YES

input

DOGEFORCES

output

NO

暴力找切割点

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
char s[MAXN],s2[MAXN]="CODEFORCES";
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	
	while(scanf("%s",s)==1)
	{
		int n=strlen(s);
		int m=strlen(s2); 
		
		if (n<m)
		{
			cout<<"NO"<<endl;
			continue;
		}
		
		int t=n-m;
		
	
		Rep(i,n-t+1)
		{
			int flag=0;
			Rep(j,i) if (s[j]!=s2[j]) flag=1;
			if (flag) continue;
			Fork(j,i+t,n-1) if (s[j]!=s2[j-t]) flag=1;
			if (flag==0)
			{
				cout<<"YES"<<endl;
				return 0;
			}
		}
		
		
		cout<<"NO"<<endl;
		
		
	}
	
	return 0;
}