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>

#define M 1046257
#define NIL (-1)
#define L 14

char H[M][L];

// 文字から数値に変換
int getChar(char ch) {
  if ( ch == 'A' ) return 1;
  else if ( ch == 'C' ) return 2;
  else if ( ch == 'G' ) return 3;
  else if ( ch == 'T' ) return 4;
  return 0;
}

// 文字列から数値へ変換して key を生成する
long long getKey(char str[]) {
  long long sum =0, p = 1, i;
  for(i=0;i<strlen(str);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[]) {
  long long key, i, h;
  key = getKey(str); // 文字列を数値に変換
  for(i=0; ;i++) {
    h = (h1(key) + i*h2(key)) %M;
    if( strcmp(H[h],str) == 0) return 1;
    else if ( strlen(H[h]) == 0 ) return 0;
  }
  return 0;
}

int insert(char str[]) {
  long long key, i, h;
  key = getKey(str); // 文字列を数値に変換
  for(i=0; ; i++) {
    h = (h1(key) + i * h2(key)) %M;
    if(strcmp(H[h],str) == 0) return 1;
    else if(strlen(H[h]) == 0) {
      strcpy(H[h],str);
      return 0;
    }
  }
  return 0;
}

int main(void) {
  int i, n;
  char str[L], com[9];
  for(i=0;i<M;i++) H[i][0] = '\0';
  scanf("%d",&n);
  for(i=0;i<n;i++) {
    scanf("%s %s",com,str); // より高速なscanfを使用
    if(com[0] == 'i') {
      insert(str);
    } else {
      if(find(str)){
        printf("yes\n");
      } else {
        printf("no\n");
      }
    }
  }
  return 0;
}