CodeForces - 538A 폭력

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). Example Input CODEWAITFORITFORCES Output YES Input BOTTOMCODER Output NO Input DECODEFORCES Output YES Input DOGEFORCES Output
NO
 
[제목]
문자열을 하나 주고 문자열을 빼서 나머지는 CODEFORCES로 만들 수 있는지 물어보기;
하나는 빼야 돼.
【해법】
포스1, 포스2로 절단점, 폭력을 기록하다.
비교적 멍청한 판법이 히히히히.
 

#include
#include
char s[105];
int cf[15];
int l;

int judge(int pos1,int pos2){
	int pos=0;
	if(l-(pos2-pos1+1)!=10)	return 0;//             
	for(int i=0;il)	break;
		if(pos==0){
			if(s[i]=='C')	cf[pos++]=1;
			else return 0;	
		}
		else if(pos==1){
			if(s[i]=='O')	cf[pos++]=1;
			else return 0;	
		}
		else if(pos==2){
			if(s[i]=='D')	cf[pos++]=1;
			else return 0;	
		}
		else if(pos==3){
			if(s[i]=='E')	cf[pos++]=1;
			else return 0;	
		}
		else if(pos==4){
			if(s[i]=='F')	cf[pos++]=1;
			else return 0;	
		}
		else if(pos==5){
			if(s[i]=='O')	cf[pos++]=1;
			else return 0;	
		}
		else if(pos==6){
			if(s[i]=='R')	cf[pos++]=1;
			else return 0;	
		}
		else if(pos==7){
			if(s[i]=='C')	cf[pos++]=1;
			else return 0;	
		}
		else if(pos==8){
			if(s[i]=='E')	cf[pos++]=1;
			else return 0;	
		}
		else if(pos==9){
			if(s[i]=='S')	return 1;
			else return 0;	
		}
	}
	return 0;
}
int main(){
	scanf("%s",s);
	int flag=0;
	l=strlen(s);
	for(int i=0;i