Problem:
Search III

Your task is to write a program of a simple dictionary which implements the following instructions:

insert str: insert a string str in to the dictionary

find str: if the distionary contains str, then print 'yes', otherwise print 'no'

Input

In the first line n, the number of instructions is given. In the following n lines, n instructions are given in the above mentioned format.

Output

Print yes or no for each find instruction in a line.

Constraints

A string consists of 'A', 'C', 'G', or 'T'

1 ≤ length of a string ≤ 12

n ≤ 1000000

Sample Input 1

5
insert A
insert T
insert C
find G
find A

Sample Output 1

no
yes

Sample Input 2

13
insert AAA
insert AAC
insert AGA
insert AGG
insert TTT
find AAA
find CCC
find CCC
insert CCC
find CCC
insert T
find TTT
find T

Sample Output 2

yes
no
no
yes
yes
yes

Notes

Template in C

Solution in C (judge verdict: Accepted)

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#include<math.h>

#define M 1000010
#define L 20
int answer[M];
char dictionary[M][L];
int getChar(char ch)
{
	if(ch=='A')
		return 1;
	if(ch=='C')
		return 2;
	if(ch=='G')
		return 3;
	if(ch=='T')
		return 4;
	return 0;
}
long long int getkey(char str[])
{
	long long int sum=0,p=1,i;
	for(i=0;i<=strlen(str)-1;i++)
	{
		sum+=p*getChar(str[i]);
		p*=5;
	}
	return sum;
}
int h1(int key)
{
	return key%M;
}
int h2(int key)
{
	return 1+(key%(M-1));
}
int find(char str[20])
{
	long long int key,i,h;
	key=getkey(str);
	for(i=0;;i++)
	{
		h=(h1(key)+i*h2(key))%M;
		if(strcmp(str,dictionary[h])==0)
			return 1;
		if(strlen(dictionary[h])==0)
			return 0;
	}
	return 0;
}
void insert(char str[20])
{
	long long int key,i,h;
	key=getkey(str);
	for(i=0;;i++)
	{
		h=(h1(key)+i*h2(key))%M;
		if(strcmp(str,dictionary[h])==0)
			return;
		if(strlen(dictionary[h])==0)
		{
			strcpy(dictionary[h],str);
			return;
		}
	}
	return;
}
int main()
{
	char str[L],com[9];
	int n,i,answern=0;
	//for(i=0;i<M;i++)
	//	dictionary[i][0]='\0';

	scanf("%d",&n);	
	for(i=1;i<=n;i++)
	{
		scanf("%s %s",com,str);
		if(strcmp(com,"insert")==0)
			insert(str);
		if(strcmp(com,"find")==0)
		{
			answern++;
			answer[answern]=find(str);
		}
	}
	for(i=1;i<=answern;i++)
	{
		if(answer[i])
			printf("yes\n");
		else
			printf("no\n");
	}
	return 0;
}