1078. Hashing (25)

最新推荐文章于 2022-10-08 00:43:42 发布

多一些不为什么的坚持

最新推荐文章于 2022-10-08 00:43:42 发布

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本文介绍了一种使用二次探测法解决哈希表中数值冲突的方法。具体地，当发生冲突时，采用二次探测（Quadratic Probing）的方式寻找下一个空闲位置，并通过实例演示了如何处理用户输入的不同正整数，最终输出这些数在经过哈希处理后的实际位置。

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 -

这题过不去的应该就是最后一个点。

Quadratic probing二次探测。之前也没接触过。

就是这个数如果已经插入进去，进行二次检测

num+0^2

num+1^2

num+2^2....

然后%msize在判断

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
int prime(int x){
	if(x==1) return 0;
	if(x==2||x==3) return 1;
	for(int i=2;i<=sqrt(x);i++) if(x%i==0) return 0;
	return 1;
}
int main(){
	int msize,n;
	cin>>msize>>n;
	int fz=msize;
	if(!prime(msize)){
		for(int i=msize+1;;i++){
			if(prime(i)){
				msize=i;
				break;
			}
		}
	}
	int hash[10001]={0};
	int v[10001]={0}; 
	int first=10000000;
	int num;
	cin>>num;
	cout<<num%msize;
	hash[num%msize]=1;
	for(int i=1;i<n;i++){
		int num;
		cin>>num;
		int index=0;
		int flag=0;
		if(hash[num%msize]==1){
			for(int j=0;j<msize;j++){
				index=(num+j*j)%msize;
				if(hash[index]==0){
				   hash[index]=1;
				   cout<<" "<<index;
				   flag=1;
				   break;	
				}
			}
			if(!flag) cout<<" "<<'-'; 
		}
		else{
			cout<<" "<<num%msize;
			hash[num%msize]=1;
		}
	}
	return 0;
}