【笨方法学PAT】1078 Hashing （25 分）

最新推荐文章于 2022-04-12 11:34:44 发布

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最新推荐文章于 2022-04-12 11:34:44 发布

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本文详细解析了一种使用二次探方插入法解决哈希表冲突的算法，介绍了如何通过调整哈希表大小为最接近的素数来优化存储效率，并实现了一段C++代码示例，展示了输入一系列正整数并输出其在哈希表中位置的过程。

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一、题目

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 -

二、题目大意

二次探方插入。

三、考点

hash

四、注意

1、探方：k = (a + j * j) % m;

2、【注意】1不是素数，否则3分过不去。

五、代码

#include<iostream>
#include<vector>
using namespace std;

bool isPrime(int n) {
	if (n == 1)
		return false;
	if (n == 2 || n == 3)
		return true;
	for (int i = 2; i*i <= n; ++i) {
		if (n%i == 0)
			return false;
	}
	return true;
}

int main() {
	//read
	int m, n;
	cin >> m >> n;

	//get prime number
	while (!isPrime(m))
		m++;

	vector<int> hash(m,-1);
	for (int i = 0; i < n; ++i) {
		int a;
		cin >> a;

		//find hash
		int k;
		bool flag = false;
		for (int j = 0; j < m; ++j) {
			k = (a + j * j) % m;
			if (hash[k] == -1) {
				hash[k] = a;
				flag = true;
				break;
			}
		}

		//judge space
		if (i != 0)
			cout << " ";

		//judge hash
		if (flag == true) 
			cout << k;
		else
			cout << "-";
	}

	system("pause");
	return 0;
}