1078. Hashing (25)

最新推荐文章于 2020-10-31 03:11:14 发布

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#hashtable #Quadratic probing

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本文介绍如何使用二次探测法解决哈希表中的冲突问题，并实现了一个简单的哈希表插入算法。具体包括如何确定哈希表的大小、二次探测的具体步骤及其实现代码。

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 -

任务是建立一个哈希表，处理冲突的方法是二次探测法（只往增加的方向探测，即根据公式 h_i=(h(key)+i*i)％m 0≤i≤m-1 ）。注意哈希表的大小规定是大于或等于给出的大小msize的最小素数。所以一开始要计算哈希表的大小，然后对于每个要插入的数，进行二次探测（最多查找m遍，一旦遇到一个位置是空的就停止循环），然后最后的位置如果能插入就插入且输出位置，不能插入就输出“-“。

代码：

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <map>
#include <cmath>
using namespace std;

bool isprime(int n)
{
	if(n<=1) return false;
	for(int i=2;i<=sqrt(n);i++)
	{
		if(n%i==0) return false;
	}
	return true;
}

int main()
{
	int msize,n;
	cin>>msize>>n;
	while(!isprime(msize))
	{
		msize++;
	}
	vector<bool>isoccupied(n,false);
	for(int i=0;i<n;i++)
	{
		int a;
		cin>>a;
		int idx1=a%msize,idx2=idx1;
		if(i!=0) cout<<" "; 
		int d=1,j=0;
		while(isoccupied[idx2]&&j<msize)
		{
			idx2=(idx1+d*d)%msize;
			d++;
			j++;
		}
		if(!isoccupied[idx2])
		{
			cout<<idx2;
			isoccupied[idx2]=true;
		}
		else
		{
			cout<<"-";
		}
	}
}