1145 Hashing - Average Search Time (25 分)

2019-05-19

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.

Sample Input:

1
2
3

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

Sample Output:

1 2	15 cannot be inserted. 2.8

作者: CHEN, Yue

单位: 浙江大学

时间限制: 200 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

要求计算平方探测哈希表的平均查询长度

分析

首先表的大小要换成比给定尺寸大的第一个素数。然后使用平方探测插入，只有正数，1,4,9这样的。查询时，计算位置，如果是该位置上是该数说明找到了，或0说明此数不在表中。

坑点：如果没找到，次数会比表长大1，这个不是很理解。

代码

#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;
int msize,n,m,t;
vector<int> table;
bool isprime(int k){
    if(k==2||k==3)
        return true;
    for(int j=2;j<=sqrt(k);j++)
        if(k%j==0)
            return false;
    return true;
}
int getprimes(int size){
    for(int i=size;;i++){
        if(isprime(i))
            return i;
    }
}
int main() {
    cin>>msize>>n>>m;
    msize=getprimes(msize);
    table.resize(msize);
    for(int i=0;i<n;i++){
        cin>>t;
        int flag=0;
        for(int j=0;j<msize;j++){
            int pos=(t+j*j)%msize;
            if(table[pos]==0){
                table[pos]=t;
                flag=1;
                break;
            }
        }
        if(flag==0)
            cout<<t<<" cannot be inserted."<<endl;
    }
    int sum=0;
    for(int i=0;i<m;i++){
        cin>>t;
        int cnt=1;
        for(int j=0;j<msize;j++){
            int pos=(t+j*j)%msize;
            if(table[pos]==t||table[pos]==0)
                break;
            cnt++;
        }
        sum+=cnt;
    }
    printf("%.1f\n",sum*1.0/m);
}