5-17 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 H(key) = key \% TSizeH(key)=key%TSizewhere TSizeTSize 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: MSizeMSize(\le 10^4≤10?4??) and NN (\le MSize≤MSize) which are the user-defined table size and the number of input numbers, respectively. Then NN 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 -

/*
 * 这题就是除留余数法插入 平方探测法解决冲突（这里只有正向） 实现散列表。
*/
#include "iostream"
#include "cmath"
using namespace std;
#define MAXSIZE 20000 /* 带进去算了下这区间有素数 10006 我也是试着取得0..0.。*/
int nextPrime(int n) {
    int i, j;
    bool flag = true;
    if (n % 2 == 0)
        n++;
    if (n == 1) /* 判断素数要注意1啊~ 在这里卡了2次- - */
        return 2;
    for (i = n; i < MAXSIZE; i+=2) {
        int k = sqrt(i);
        for (j = 2; j<=k ; j++)
            if (!(i%j))
                break;
        if (j > k)
            return i;
    }
    return i;
}

int val[10001];
void find(int a[],int m,int n) {
    int i, j;
    for (i = 0; i < m; i++) {
        for (j = 0; j < n; j++) {
            int pos = (val[i]%n + j*j) % n;
            if (a[pos] == val[i])
            {
                if (i == 0)
                    cout << pos;
                else
                    cout << " " << pos;
                break;
            }
        }
        if (j == n)
            if (i == 0) {
                cout << "-";
            }
            else {
                cout << " " << "-";
            }
    }
}

void insert(int n,int m,int a[]) {
    for (int i = 0; i < m; i++) {
        cin >> val[i];
        for (int j = 0; j < n; j++) {
            int pos = (val[i]%n + j*j) % n;
            if (!a[pos]) {
                a[pos] = val[i];
                break;
            }
        }
    }
}
int main() {
    int n, m;
    int a[20001];
    cin >> n >> m;
    n = nextPrime(n);
    for (int i = 0; i < n; i++)
        a[i] = 0;
    insert(n, m, a); /* 映射 */
    find(a, m, n);
    cout << endl;
    return 0;
}