【PAT】【Advanced Level】1078. Hashing (25)

最新推荐文章于 2024-05-05 11:45:29 发布

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最新推荐文章于 2024-05-05 11:45:29 发布

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本文介绍了一道关于哈希表的编程题，重点在于使用二次探测法解决哈希冲突。文章给出了完整的代码实现，并解释了如何确定合适的哈希表大小及处理碰撞的方法。

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1078. Hashing (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 -

原题链接：

https://www.patest.cn/contests/pat-a-practise/1078

https://www.nowcoder.com/pat/5/problem/4308

题目大意：

哈希，冲突用二次探测法解决

Quadratic probing（二次探测法）

吃了英语渣的亏。。

思路：

普通的哈希加上二次探测法的冲突处理

CODE：

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


map<int,int> ma;


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


int main()
{
	int pri,n;
	cin>>pri>>n;
	while (!prime(pri)){pri++;}
	//cout<<pri<<endl;
	int t;
	for (int i=0;i<n;i++)
	{
		cin>>t;
		int tt=t%pri;
		if (ma[tt]==0)
		{
			cout<<tt;
			ma[tt]=1;
		}
		else
		{
			int fl=0;
			for (int j=1;j<=pri;j++)
			{
				tt=(t+j*j)%pri;
				if (ma[tt]==0)
				{
					cout<<tt;
					ma[tt]=1;
					fl=1;
					break;
				}
			}
			if (fl==0)
				cout<<"-";
		}
		if (i!=n-1)	cout<<" ";
	}
	return 0;
}