7-17 Hashing（25 分）

最新推荐文章于 2021-12-28 14:20:03 发布

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最新推荐文章于 2021-12-28 14:20:03 发布

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文章标签： java

原文链接：http://www.cnblogs.com/A-Little-Nut/p/8159222.html

本文介绍了一种使用哈希表存储唯一正整数的方法，并通过平方探测法解决冲突。当用户定义的表大小不是素数时，会重新定义为下一个素数。输入包括表的大小和待插入的整数，输出则是每个整数的位置。

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Hashing

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 (≤10^4)

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 -

分析

从本题给的样例输出可以看出本题使用平方探测法

代码如下

#include<iostream>
#include<math.h>
using namespace std;
int IsPrime(int n)
{
    if(n <= 1) return 0;
    for(int i=2;i<=sqrt(n);i++)
    if(n%i==0) 
    return 0;   
    return 1;
}
int NextPrime(int n)
{
    while(!IsPrime(n))n++;
    return n;
}
int main(){
    int m,n,tblsize,t,p;
    cin>>m>>n;
    tblsize=NextPrime(m);
        int pos[tblsize];
        for(int i=0;i<tblsize;i++)
        pos[i]=0;
    for(int i=0;i<n;i++){
        cin>>t; int j,flag=0;
        for(j=0;j<tblsize;j++)
        {
            p=(t%tblsize+j*j)%tblsize;
            if(pos[p]==0){
                pos[p]=1;
                flag=1;
                break;
            }
        }
        if(!i) flag>0?cout<<p:cout<<"-";
        else flag>0?cout<<" "<<p:cout<<" -";
    }
    return 0;
}

转载于:https://www.cnblogs.com/A-Little-Nut/p/8159222.html