1078 Hashing (25分)(欧拉筛素数打表 + hash平方探测)

1078 Hashing (25分)

 

The task of this problem is simple: insert a sequence of distinct positive integers into a hash table, and output the positions of the input numbers. 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 two positive numbers: MSize (≤) and N (≤) which are the user-defined table size and the number of input numbers, respectively. Then N distinct positive integers are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print the corresponding positions (index starts from 0) of the input numbers in one line. All the numbers in a line are separated by a space, and there must be no extra space at the end of the line. In case it is impossible to insert the number, print "-" instead.

Sample Input:

4 4
10 6 4 15

 

Sample Output:

0 1 4 -



首先讲一下关于哈希表里面平方探测的小知识。


假设哈希函数h(x) = x % prime;
我们知道哈希有很多解决冲突的办法，其中一种就是平方探测。
平方探测就是，用(0, prime)的平方 + x继续进行哈希，直到找到一个不冲突的位置。
为什么是上限是prime - 1呢？
因为如果平放到prime时，(prime * prime + x) % prime == ((prime % prime) * (prime % prime) + x % prime) % prime == x % prime，往下你会发现都是重复的，因此上限是prime - 1。
还有记得....记得prime一定要是一个素数。

#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 1e5 + 5;
bool vis[maxn];
int prime_list[maxn], cnt;

void init() {
    for(int i = 2; i < maxn; i ++) {
        if(!vis[i]) {
            prime_list[cnt ++] = i;
        }
        for(int j = 0; j < cnt; j ++) {
            if(i * prime_list[j] > maxn) break;
            vis[i * prime_list[j]] = true;
            if(i % prime_list[j] == 0) break;
        }
    }
}

bool _hash[maxn];

int main() {
    init();
    int n, prime, val;
    scanf("%d %d", &prime, &n);
    for(int i = 0; i < maxn; i ++) {
        if(prime_list[i] >= prime) {
            prime = prime_list[i];
            break;
        }
    }
    bool flag;
    for(int i = 0; i < n; i ++) {
        scanf("%d", &val);
        flag = false;
        for(int j = 0; j < prime; j ++) {
            if(!_hash[(j * j + val) % prime]) {
                _hash[(j * j + val) % prime] = true;
                printf("%d%c", (j * j + val) % prime, i == n - 1 ? '
' : ' ');
                flag = true;
                break;
            }
        }
        if(!flag) printf("-%c", i == n - 1 ? '
' : ' ');
    }
    return 0;
}