PAT Grade --A1078 Hashing

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 $( where$

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:

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 -

Element values given by the remainder of the length of the table as the insertion position in the hash table, if a conflict occurs, using the square method to solve the probe. The specific process is square probe method, assume a value of a given element a, the table length is M, the insertion position of a% M, assuming a% M existing element position, i.e. collision occurs, look

(. 1 + A ^ 2)% M, (A-2 ^. 1)% M, (A + 2 ^ 2)% M, (A-2 ^ 2)% M, no no , (A + M ^ 2) % M, (aM ^ 2) % M until you find the position of an insertion, or when found (a + M ^ 2)% M, (aM ^ 2)% M is still not inserted into the element insertion fails.

 1 #include <iostream>
 2 #include <vector>
 3 #include <cmath>
 4 using namespace std;
 5 int M, N;
 6 bool isPrime(int num)
 7 {
 8     if (num <= 3)
 9         return num > 1;
10     if (num % 6 != 1 && num % 6 != 5)
11         return false;
12     int s = (int)sqrt(num);
13     for (int i = 5; i <= s; ++i)
14         if (num % i == 0)
15             return false;
16     return true;
17 }
18 int main()
19 {
20     cin >> M >> N;
21     while (!isPrime(M++));//找到素数
22     M--;
23     vector<int>table(M, 0);
24     int num, index;
25     for (int i = 0; i < N; ++i)
26     {
27         cin >> num;
28         index = num % M;
29         if (table[index] > 0)//存在冲突
30         {
31             for (int i = 1; i <= M; ++i)
32             {
33                 index = (num + i * i) % M;
34                 if (table[index] == 0)
35                     break;        
36             }
37         }
38         if(table[index] == 0)//不存在冲突
39         {
40             table[index]++;
41             cout << index;
42         }
43         else
44             cout << "-";
45         if (i < N - 1)
46             cout << " ";
47     }
48     return 0;
49 }