PAT 1078. Hashing (25)

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本文探讨了如何使用Hash表及二次探测解决碰撞问题，并详细解释了当表大小非质数时的处理方法。通过具体实例展示了算法实现过程。

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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 -

题目大意

1.要求是插入一个正数到hash表，输出他的位置。
2.如果hash表的大小不是质数，把它换成大于它的最小的质数。
3.不能插入某个数字，输出-。
4.二次探测，key+i^2。

解题思路

1.一直在想二次探测怎么退出循环，还以为它要一直加到无穷大呢卧槽，谁知道只加到n 就行了，意思是对于key+i^2，找不到的话就让它从0找到插入元素的个数n就行了。
2.注意1不是质素，是的话分解质因数就不唯一了。

AC代码

#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
int visited[10001];
int Msize, n;
bool isPrime(int a){
    if (a==1)
    {
        return false;
    }
    for (int i = 2; i <= sqrt(a); i++)
    {
        if (a%i==0)
        {
            return false;
        }
    }
    return true;
}
int main()
{
    cin >> Msize >> n;
    while (!isPrime(Msize))
    {
        Msize++;
    }
    //cout << Msize << endl;
    int l;
    for (int i = 0; i < n; i++)
    {
        cin >> l;
        //平方探测是从0到放入元素的个数n为止？
        int j=0;
        //那么就从0到n遍历，如果那个位置是空的就直接退出循环就好了，
        //如果没有空就继续找
        for (j = 0; j < n; j++)
        {
            int tem = (l + j*j) % Msize;
            if (visited[tem] == 0){
                visited[tem] = 1;
                if (i == 0)
                {
                    cout << tem;
                }
                else{
                    cout << " " << tem;
                }
                break;
            }
        }
        //如果找不到输出-
        if (j==n)
        {
            ;
            if (i==0)
            {
                cout << "-";
            }
            else{
                cout << " -";
            }
        }

    }
    cout << endl;
    return 0;
}