PAT 1145 Hashing - Average Search Time（25 分）C++ [自己开发的博客网站，欢迎访问](www.weiboke.online) www.weiboke.o

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哈希冲突，并计算平均查找时间。文章提供了详细的算法实现步骤及代码示例。

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#1145 Hashing - Average Search Time（25 分）
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

##Approach
1.这道题要求模拟哈希表的过程，然后计算平均查找次数为多少，然后这道题有点迷，不知道为什么没查找到也要增加一次次数，要不然只能过3个测试点。这里主要的知识点就是题目要求的是平方探测法（Quadratic probing）而且还只用正数。
2.平方探测法（Quadratic probing）：当表中下标为H(key)的位置被占用的时候，将按顺序检查表中的位置：H(key)+1^2、H(key)-12、H(key)+2^2、H(key)-22、H(key)+3^2、H(key)-32、H(key)+4^2、H(key)-42…。如果检查过程中H(key)+k^{2超过了表长TSIZE，那么就把(H(key)+k}2)%TSIZE。如果H(key)-k^{2<0，则((H(key)-k}2)%TSIZE+TSIZE)%TSIZE。

Code

void make_prime(vector<bool>& prime, int N) {
	prime[0] = false;
	prime[1] = false;
	for (int i = 2; i < N; i++) {
		if (prime[i]) {
			for (int j = i*i; j < N; j += i) {
				prime[j] = false;
			}
		}
	}
}
bool inserthashtable(vector<int>&hash,int element,int &msize) {
	for (int i = 0; i < msize; i++) {
		int key = (element%msize + i*i) % msize;
		if (hash[key] == -1) {
			hash[key] = element;
			return true;
		}
	}
	return false;
}
void searchhashtable(vector<int>&hash, int element, int &msize, int &cnt) {
	for (int i = 0; i < msize; i++) {
		int key = (element%msize + i*i) % msize;
		cnt++;
		if (hash[key] == element || hash[key] == -1) {
			return;
		}
	}
	cnt++;
	return;
}
int main() {
	vector<bool> prime(10015, true);
	make_prime(prime, 10015);
	int msize, n, m;
	scanf("%d %d %d", &msize, &n, &m);
	while (!prime[msize])msize++;
	vector<int>hash(msize, -1);
	for (int i = 0; i < n; i++) {
		int element;
		scanf("%d", &element);
		if (!inserthashtable(hash, element,msize)) {
			printf("%d cannot be inserted.\n", element);
		}
	}
	int cnt = 0;
	for (int i = 0; i < m; i++) {
		int element;
		scanf("%d", &element);
		searchhashtable(hash, element, msize, cnt);
	}
	printf("%.1f\n", cnt*1.0 / m);
	return 0;
}