ALDS1_4_C Dictionary(哈希）

Discussion

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

题意

实现一个简易的字典，能insert（插入） 和 find（查找）

思路

数据范围大，可能查询很多次，所以用HASH来储存，插入和查找的时间复杂度都是O（1）。

代码

#include<stdio.h>#include<string.h>const int L = 14;const int M = 1046527;char H[M][L]; /* Hash Table */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;}/* convert a string into an integer value */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 = 0;    key = getKey(str);    while(1)    {        int h = ( h1(key) + i*h2(key) ) % M;        if(strcmp( H[h],str) == 0 ) return 1;        else if(strlen(H[h]) == 0) return 0;        i++;     }    return 0;}void insert(char str[]){    long long i = 0,key;    key = getKey(str);    while(1)    {        int h = ( h1(key) + i*h2(key) ) % M;        if(strcmp(H[h],str) == 0 ) return ;        else if(strlen(H[h]) == 0 )        {            strcpy(H[h],str);            return ;        }        i++;    }    return ;}int main(){    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);    if ( com[0] == 'i' ){        insert(str);    } else {        if (find(str)){        printf("yes\n");        } else {        printf("no\n");        }    }    }    return 0;}