【笨方法学PAT】1145 Hashing - Average Search Time （25 分）

一、题目

The task of this problem is simple: insert a sequence of distinct positive integers into a hash table first. Then try to find another sequence of integer keys from the table and output the average search time (the number of comparisons made to find whether or not the key is in the table). 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 3 positive numbers: MSize, N, and M, which are the user-defined table size, the number of input numbers, and the number of keys to be found, respectively. All the three numbers are no more than 10​4​​. Then N distinct positive integers are given in the next line, followed by M positive integer keys in the next line. All the numbers in a line are separated by a space and are no more than 10​5​​.

Output Specification:

For each test case, in case it is impossible to insert some number, print in a line X cannot be inserted. where X is the input number. Finally print in a line the average search time for all the M keys, accurate up to 1 decimal place.

Sample Input:

4 5 4
10 6 4 15 11
11 4 15 2

Sample Output:

15 cannot be inserted.
2.8

二、题目大意

给定一组数据，要求插入长度为MSize的哈希表中。 
再给定一组数据，查找各元素在哈希表中的位置。输出总查找次数的平均值，保留一位小数。

三、考点

Hash

四、注意

1、插入：Hi=(key+j*j)%TSize(开放地址发——二次方探测再散列)，其中 j>=0&&j<TSize;

2、查找：Hi=(key+j*j)%TSize，其中 j>=0&&j<=TSize；

3、查找的时候注意判断位置是否占用或者空置。

五、代码

#include<iostream>
#include<vector>
using namespace std;
bool isPrime(int n) {
	if (n == 2 || n == 3)
		return true;
	for (int i = 2; i*i <= n; ++i) {
		if (n%i == 0)
			return false;
	}
	return true;
}
int main() {
	//read
	int t, n, m;
	cin >> t >> n >> m;

	//find the prime size
	while (!isPrime(t))
		t++;
	vector<int> vec(t, -1);

	//insert numbers
	for (int i = 0; i < n; ++i) {
		int key;
		cin >> key;
		
		//^2
		bool flag = false;
		for (int j = 0; j < t; ++j) {
			int h = (key + j * j) % t;
			if (vec[h] == -1) {
				vec[h] = key;
				flag = true;
				break;
			}	
		}

		//can not insert
		if(flag==false)
			printf("%d cannot be inserted.\n", key);
	}

	//find numbers
	int num = 0;
	for (int i = 0; i < m; ++i) {
		int key;
		cin >> key;

		//^2
		for (int j = 0; j <= t; ++j) {
			int h = (key + j * j) % t;
			num++;
			if (vec[h] == key||vec[h]==-1) {
				break;
			}
		}
	}

	//output
	printf("%.1f", num*1.0 / m);

	system("pause");
	return 0;
}