codeforces 538 A.Cutting Banner

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原创最新推荐文章于 2026-05-10 09:30:48 发布 1.1k 阅读 AI 写同款· GEO 优化 ›

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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 outTEM), '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

题目链接：http://codeforces.com/contest/538/problem/A

题目大意：给一个字符串，求切下一段后剩余部分是否等于“CODEFORCES”，能输出“YES”，不能输出"NO"。常见的字符串处理问题。用substr()函数处理即可。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;
int main()
{
	int n,p=0;
	string a;
	cin>>a;
	n=a.length();
	for(int i=0;i<n;i++)
	{
		for(int len1=1;len1<n-i+1;len1++)
		{
			string b=a.substr(i,len1);//b为切后的前一段，c为切后的后一段
			for(int j=i+len1;j<n;j++)
			{
				for(int len2=1;len2<n-j+1;len2++)
				{
					string c=a.substr(j,len2);
					if(b+c=="CODEFORCES"&&(i==0&&j+len2==10||j+len2==n&&i==n-10||i==0&&j+len2==n))
					{//因为只能切掉一段，所以剩余两段连着在串首或者连着在串尾或者分别在两端
						p=1; 
						printf("YES\n");
						break;
					}
				}
				if(p)
					break;
			}
			if(p)break;
		}
		if(p)break;
	}
	if(!p)
		printf("NO\n");
	return 0;
}

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