PAT 1145 Hashing - Average Search Time （25 分）

最新推荐文章于 2019-10-07 19:14:05 发布

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最新推荐文章于 2019-10-07 19:14:05 发布

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分类专栏： PAT 文章标签： PAT Hash

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本文详细解析PAT1145 Hashing-AverageSearchTime问题，涉及hash表构建、冲突解决策略及平均查找时间计算。采用除留余数法和平方探测法处理冲突，特别关注表大小为素数的要求。

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PAT 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 10⁴. 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 10⁵ .

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.

Simple Input:

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

Simple Output:

15 cannot be inserted.
2.8

题目大意

题目意思很简单，建立一个hash表，采用除留余数法确定hash函数，采用平方探测法（仅考虑正向递增序列）来作为冲突解决策略。需要注意的是，题目给出了hash表的初始大小MSize，如果MSize是素数就不用管了，如果不是，你需要找到比它大的最近的素数。题目先给出N个数让你建表，随后给出M个数让你查询，最后需要先打印不能插入在表中的数，再打印平均搜索时间。

解题思路

这里主要是平均查找次数的计算，包含两种情况，一种在MSize-1范围内能找到，对应于i<MSize；一种是在MSize-1范围内找不到，对应于i==MSize；这里需要注意的是，实际查找次数应该为i+1。当判断为i==MSize时，要把对应的key记录下来。

C++实现

#include <iostream>
#include <cstdio>
#include <vector>
#include <cmath>

using namespace std;

bool isPrime(int X)
{
	bool flag=true;
	int i;
	if(X==1){
		flag=false;
	}else if(X%2==0){
		flag=false;
	}else{
		for(i=3;i<=sqrt(X);i++){
			if(X%i==0){
				flag=false;
				break;
			}
		}
	}
	return flag;
}

int NextPrime(int MS)
{
	int MSize;
	if(isPrime(MS)){
		MSize=MS;
	}else{
		if(MS==1){
			MSize=2;
		}else{
			MS=MS%2==0 ? MS+1:MS+2; //MS为奇数 
			while(!isPrime(MS)){
				MS+=2;
			}
			MSize=MS;
		}
	}
	return MSize;
}

void Insert(int Hash[],int MS,int Key)
{
	int i,nk;
	for(i=0;i<MS;i++){
		nk=(Key%MS+i*i)%MS;
		if(Hash[nk]==0){
			Hash[nk]=Key;
			break;
		}
	}
}

int Find(int Hash[],int MS,int Key)
{
	int i,nk;
	for(i=0;i<MS;i++){
		nk=(Key%MS+i*i)%MS;
		if(Hash[nk]==Key || Hash[nk]==0){
			return i;   //找到了 
			break;
		}
	}
	return i;  //找不到，此时i==MS 
}

int main()
{
	int MS,N,M;
	cin>>MS>>N>>M;
	MS=NextPrime(MS);
	int Hash[MS]={0};
	int key;
	while(N--){
		cin>>key;
		Insert(Hash,MS,key);
	}
	float AST=0;
	vector<int> CBI;
	int cnt,i;
	for(i=0;i<M;i++){
		cin>>key;
		cnt=Find(Hash,MS,key);
		AST+=cnt+1;
		if(cnt==MS){
			CBI.push_back(key);
		}
	}
	if(CBI.size()>0){
		for(i=0;i<CBI.size();i++){
			printf("%d ",CBI[i]);
		}
		printf("cannot be inserted.\n");
	}
	
	printf("%.01f",AST/M);
}