PAT 甲级 1145 Hashing - Average Search Time (25 分)（读不懂题，也没听说过平方探测法解决哈希冲突。。。感觉题目也有点问题）

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 ( 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 1. 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 1.

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

题意：

给定一个序列，用平方探测法解决哈希冲突，然后给出m个数字，如果这个数字不能够被插入就输出”X cannot be inserted.”，然后输出这m个数字的平均查找时间

题解：

找到大于tsize的最小的素数为真正的tsize，然后建立一个tsize长度的数组。首先用平方探测法插入数字a，每次pos = (a + j * j) % tsize，j是从0～tsize-1的数字，如果当前位置可以插入就将a赋值给v[pos]，如果一次都没有能够插入成功就输出”X cannot be inserted.”。其次计算平均查找时间，每次计算pos = (a + j * j) % tsize，其中j <= tsize，如果v[pos]处正是a则查找到了，则退出循环，如果v[pos]处不存在数字表示没查找到，那么也要退出循环。每次查找的时候，退出循环之前的j就是这个数字的查找长度。最后ans除以m得到平均查找时间然后输出～

而我看不懂题，也没听说过Quadratic probing平方探测法解决哈希冲突，看来要多扩充知识点了。

　　哈希函数构造方法：H(key) = key % TSize （除留余数法）
　　处理冲突方法：Hi = (H(key) + di) % TSize (开放地址发——二次方探测再散列)

　　其中di为 1*1 , -1*1 , 2*2 , -2*2 , ··· k*k , -k*k (k <= MSize-1)

　　题目中提到 with positive increments only 所以我们只需要考虑正增量即可。

看了很多人本题的题解，我发现有些说要前一次 [0，tsize)，但后面计数是要  [0,tsize]，虽然这样25分也都拿到了，但是，a%t = (a+t*t)%t 不是应该相等的吗？如果相等，又为什么计数时的那个循环里把等号去了就过不了了呢？网上有些的确时两边都统一[0，tsize)，但是在计数的时候，没有找到的要多加一次。到底两种思考方式哪个更合理呢？

AC代码：

#include<bits/stdc++.h>
using namespace std;
bool prime(int x){
    if(x<=) return false;
    for(int j=;j*j<=x;j++){
        if(x%j==) return false;
    }
    return true;
}
int main(){
    int t,m,n,x,f;
    cin>>t>>m>>n;
    while(!prime(t)) t++;
    vector<int> a(t);
    for(int i=;i<=m;i++){
        cin>>x;
        f=;
        for(int j=;j<t;j++){//j是[0,t)
            int y=(x+j*j)%t;
            if(a[y]==||a[y]==x){
                a[y]=x;
                f=;
                break;
            }
        }
        if(!f) printf("%d cannot be inserted.\n", x);
    }
    f=;
    for(int i=;i<=n;i++){
        int x;
        cin>>x;
        for(int j=;j<=t;j++){//j是[0,t]
            f++;
            int y=(x+j*j)%t;
            if(a[y]==||a[y]==x){
                break;
            }
        }
    }
    printf("%.1f",f*1.0/n);
    return ;
} 

另一种正确代码：

#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
bool isPrime(int num) {
    if (num < ) return false;
    for (int i = ; i *i<=num; i++) {
        if (num % i == ) return false;
    }
    return true;
}

int msize, n, m, a, table[];
int main() {
    memset(table, -, sizeof(table));
    scanf("%d%d%d", &msize, &n, &m);

    while (isPrime(msize) == false) msize++;

    for (int i = ; i < n; i++) {
        scanf("%d", &a);

        bool founded = false;
        for (int j = ; j < msize; j++) {
            int d = j * j;
            int tid = (a + d) % msize;
            if (table[tid] == -) {
                founded = true;
                table[tid] = a;
                break;
            }
        }
        if (founded == false) {
            printf("%d cannot be inserted.\n", a);
        }
    }
    int tot = ;

    for (int i = ; i < m; i++) {
        scanf("%d", &a);
        int t = ;
        bool founded = false;
        for (int j = ; j < msize; j++) {//这边是j从0-msize
            tot++;
            int d = j * j;
            int tid = (a + d) % msize;
            if (table[tid] == a || table[tid] == -) { // 找到或者不存在
                founded = true;
                break;
            }
        }
        if(founded ==false) {//没有找到要多加一次
            tot++;
        }
    }

    printf("%.1f\n", tot*1.0/m);

    return ;
}