Dictionary Aizu - ALDS1_4_C

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

code

/*
                                ^....0
                               ^ .1 ^1^
                               ..     01
                              1.^     1.0
                             ^ 1  ^    ^0.1
                             1 ^        ^..^
                             0.           ^ 0^
                             .0            1 .^
                             .1             ^0 .........001^
                             .1               1. .111100....01^
                             00             ^   11^        ^1. .1^
                             1.^                              ^0  0^
                               .^                                 ^0..1
                               .1                                   1..^
                             1 .0                                     ^  ^
                              00.                                     ^^0.^
                              ^ 0                                     ^^110.^
                          0   0 ^                                     ^^^10.01
                   ^^     10  1 1                                      ^^^1110.1
                   01     10  1.1                                      ^^^1111110
                   010    01  ^^                                        ^^^1111^1.^           ^^^
                   10  10^ 0^ 1                                            ^^111^^^0.1^       1....^
                    11     0                                               ^^11^^^ 0..  ....1^   ^ ^
                    1.     0^                                               ^11^^^ ^ 1 111^     ^ 0.
                   10   00 11                                               ^^^^^   1 0           1.
                   0^  ^0  ^0                                                ^^^^    0            0.
                   0^  1.0  .^                                               ^^^^    1 1          .0
                   ^.^  ^^  0^                             ^1                ^^^^     0.         ^.1
                   1 ^      11                             1.                ^^^     ^ ^        ..^
                  ^..^      ^1                             ^.^               ^^^       .0       ^.0
                  0..^      ^0                              01               ^^^       ..      0..^
                 1 ..        .1                             ^.^              ^^^       1 ^  ^0001
                ^  1.        00                              0.             ^^^        ^.0 ^.1
                . 0^.        ^.^                             ^.^            ^^^         ..0.0
               1 .^^.         .^                  1001        ^^            ^^^         . 1^
               . ^ ^.         11                0.    1         ^           ^^          0.
                0  ^.          0              ^0       1                   ^^^          0.
              0.^  1.          0^             0       .1                   ^^^          ..
              .1   1.          00            .        .1                  ^^^           ..
             1      1.         ^.           0         .^                  ^^            ..
             0.     1.          .^          .         0                                  .
             .1     1.          01          .        .                                 ^ 0
            ^.^     00          ^0          1.       ^                                 1 1
            .0      00           .            ^^^^^^                                   .
            .^      00           01                                                    ..
           1.       00           10                                                   1 ^
          ^.1       00           ^.                                            ^^^    .1
          ..        00            .1                                        1..01    ..
         1.1         00           1.                                       ..^      10
        ^ 1^         00           ^.1                                      0 1      1
        .1           00            00                                       ^  1   ^
         .           00            ^.^                                        10^  ^^
       1.1           00             00                                              10^
       ..^           1.             ^.                                               1.
      0 1            ^.              00                 00                            .^
        ^            ^.              ^ 1                00   ^0000^     ^               01
     1 0             ^.               00.0^              ^00000   1.00.1              11
     . 1              0               1^^0.01                      ^^^                01
      .^              ^                1   1^^                                       ^.^
    1 1                                                                              0.
    ..                                                                              1 ^
     1                                                                               1
   ^ ^                                                                             .0
   1                                                                             ^ 1
   ..                                                          1.1            ^0.0
  ^ 0                                                           1..01^^100000..0^
  1 1                                                            ^ 1 ^^1111^ ^^
  0 ^                                                             ^ 1      1000^
  .1                                                               ^.^     .   00
  ..                                                                1.1    0.   0
  1.                                                                  .    1.   .^
  1.                                                                 1    1.   ^0
 ^ .                                                                 ^.1 00    01
 ^.0                                                                  001.     .^
 */
// Virtual_Judge —— Dictionary Aizu - ALDS1_4_C.cpp created by VB_KoKing on 2019-05-02:11.
/* Procedural objectives:

 Variables required by the program:

 Procedural thinking:

 Functions required by the program:

*/
/* My dear Max said:
"I like you,
So the first bunch of sunshine I saw in the morning is you,
The first gentle breeze that passed through my ear is you,
The first star I see is also you.
The world I see is all your shadow."

FIGHTING FOR OUR FUTURE!!!
*/
#include <iostream>
#include <cstring>
#include <cstdio>

#define M 1046527
#define NIL -1
#define L 14

using namespace std;

char H[M][L];

//将字符转换为数值
int get_char(char ch) {
    switch (ch) {
        case 'A':
            return 1;
        case 'C':
            return 2;
        case 'G':
            return 3;
        case 'T':
            return 4;
        default:
            return 0;
    }
}

//将字符串转换为数值并生成key
long long get_key(char str[]) {
    long long sum = 0, p = 1;
    for (int i = 0; i < strlen(str); i++) {
        sum += p * (get_char(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 = get_key(str), h;
    for (int 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;
    }
}

int insert(char str[]) {
    long long key = get_key(str), h;
    for (int 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;
        }
    }
}

int main() {
    int n;
    scanf("%d",&n);
    char str[L], com[9];
    for (int i = 0; i < M; i++) H[i][0] = '\0';
    for (int i = 0; i < n; i++) {
        scanf("%s %s",com,str);
        if (com[0] == 'i') insert(str);
        else {
            if (find(str)) printf("yes\n");
            else printf("no\n");
        }
    }
    return 0;
}
posted @ 2019-05-02 14:24  AlexKing007  阅读(108)  评论(0编辑  收藏  举报