PAT A1078.Hashing

最新推荐文章于 2023-03-12 23:59:02 发布

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本文介绍了一道关于哈希表的编程题，重点在于使用平方探测法解决冲突，并确保哈希表大小为质数。文章提供了详细的解题思路及实现代码。

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 -

/**********************侯尼玛**********************/

一道简单的题目，就是做个平方探测法的哈希表，并且其中的平方探测法只要求正数增量。

唯一要注意的一点是，题目要求哈希表容量必须是大于等于给出容量的最小质数。而1不是质数……这点在看了牛客网的测试例以后才意识到。

附上代码：

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

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

int main()
{
    int N=0,Size=0;
    while(cin>>Size)
    {
        if(Size == 1) Size++;
        cin>>N;
        while(!IsPrime(Size))
            Size++;
        //cout<<Size<<endl;
        int *Hash = new int [Size]();
        int temp = 0;
        for(int i=0;i<N;i++)
        {
            cin>>temp;
            int post= temp%Size;
            if(Hash[post]==0)
            {
                Hash[post]=1;
                cout<<post;
            }
            else
            {
                int k=1;
                int NewPost = (post+k*k)%Size;
                while(Hash[NewPost]!=0 && k<Size)
                {
                    k++;
                    NewPost=(post + k*k)%Size;
                }
                if(k<Size)
                {
                    Hash[NewPost]=1;
                    cout<<NewPost;
                }
                else cout<<'-';
            }
            if(i<N-1) cout<<' ';
        }
    }
    return 0;
}