(PAT 1078) Hashing (哈希表冲突处理)

柔光的暖阳◎ 2022-03-18 02:22 210阅读 0赞

The task of this problem is simple: insert a sequence of distinct positive integers into a hash table, and output the positions of the input numbers. The hash function is defined to be H(key)=key%TSize where TSize is the maximum size of the hash table. Quadratic probing (with positive increments only) is used to solve the collisions.

Note that the table size is better to be prime. If the maximum size given by the user is not prime, you must re-define the table size to be the smallest prime number which is larger than the size given by the user.

### Input Specification: ###

Each input file contains one test case. For each case, the first line contains two positive numbers: MSize (≤104) and N (≤MSize) which are the user-defined table size and the number of input numbers, respectively. Then N distinct positive integers are given in the next line. All the numbers in a line are separated by a space.

### Output Specification: ###

For each test case, print the corresponding positions (index starts from 0) of the input numbers in one line. All the numbers in a line are separated by a space, and there must be no extra space at the end of the line. In case it is impossible to insert the number, print "-" instead.

### Sample Input: ###

4 4
    10 6 4 15

### Sample Output: ###

0 1 4 -

解题思路:

哈希表冲突处理

首先判断输入的最大容量是否是素数,如果不是,就取第一个比输入值大的素数作为容量

然后依次把值按照H(key)=key%TSize插入

如果发生冲突(该位置已经存在元素),那么就按照二次探查法寻找位置

二次探查法:假设最大容量为5,插入元素为15,设步长初始值为1,然后按照下面的步骤寻找插入位置:

H = (15 + 1\*1)%5;

H = (15 + 2\*2)%5;

H = (15 + 3\*3)%5;

H = (15 + 4\*4)%5;

如果步长大于等于最大容量5,那么就无法插入,输出 -

#include <iostream>
    #include <algorithm>
    #include <math.h>
    using namespace std;
    const int MAXN = 10010;
    int M, N;
    bool thashTable[MAXN];
    int tnumbers[MAXN];
    
    bool isPrime(int num) {
    	if (num <= 1) return false;
    	for (long long i = 2; i*i <= num; ++i) {
    		if (num % i == 0) return false;
    	}
    	return true;
    }
    
    int main() {
    	fill(thashTable, thashTable + MAXN, false);
    	cin >> M >> N;
    	for (int i = 0; i < N; ++i) {
    		cin >> tnumbers[i];
    	}
    	while (!isPrime(M)) {  //寻找最小的素数
    		M++;
    	}
    	for (int i = 0; i < N; ++i) {
    		int cposition = tnumbers[i] % M;
    		if (thashTable[cposition] == false) {
    			thashTable[cposition] = true;
    			cout << cposition;
    		}
    		else {
    			int binaryindex = 1;
    			int newpostion;
    			while (true) {
    				newpostion = (tnumbers[i] + binaryindex * binaryindex) % M;  //二次线性查找
    				if (thashTable[newpostion] == false) {
    					thashTable[newpostion] = true;
    					cout << newpostion;
    					break;
    				}
    				else {
    					binaryindex++;
    				}
    				if (binaryindex >= M) {
    					cout << "-";
    					break;
    				}
    			}
    		}
    		if (i < N - 1) cout << " ";
    	}
    
    	return 0;
    }