PAT_A1078#Hashing

Source:

PAT A1078 Hashing (25 分)

Description:

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 -

Keys:

  • 素数（Prime）
  • 散列（Hash）

Attention:

  • 二次探测法（平方探测法）H（key）= (key+d)%m，其中d= 正负1^2,2^2,...,k^2，k<=m
  • 对于数据较大的题目，可以提前打印素数表，降低时间复杂度

Code:

 1 /*
 2 Data: 2019-05-13 20:16:33
 3 Problem: PAT_A1078#Hashing
 4 AC: 32:51
 5 
 6 题目大意：
 7 哈希表中插入一些不同的正整数，输出其插入的位置
 8 哈希函数：H(key) = key%T，T为不小于MAX_SIZE的最小Prime
 9 冲突处理：二次探测法
10 
11 输入：给出M表长，N组输入
12 输出：给出插入位置,从0开始，无法插入打印“-”
13 */
14 #include<cstdio>
15 #include<vector>
16 #include<algorithm>
17 using namespace std;
18 const int M=1e4+10;
19 int isPrime[M];
20 
21 void Euler()
22 {
23     fill(isPrime,isPrime+M,1);
24     isPrime[0]=0;
25     isPrime[1]=0;
26     vector<int> prime;
27     for(int i=2; i<M; i++)
28     {
29         if(isPrime[i])
30             prime.push_back(i);
31         for(int j=0; j<prime.size(); j++)
32         {
33             if(i*prime[j] > M)
34                 break;
35             isPrime[i*prime[j]]=0;
36             if(i%prime[j]==0)
37                 break;
38         }
39     }
40 }
41 
42 int main()
43 {
44 #ifdef  ONLINE_JUDGE
45 #else
46     freopen("Test.txt", "r", stdin);
47 #endif // ONLINE_JUDGE
48 
49     Euler();
50     int n,m,x;
51     scanf("%d%d",&m,&n);
52     while(!isPrime[m])
53         m++;
54     int h[M]={0};
55     for(int i=0; i<n; i++)
56     {
57         scanf("%d", &x);
58         if(i!=0)printf(" ");
59         int k=0;
60         while(h[(x+k*k)%m]==1 && k<m)
61             k++;
62         if(h[(x+k*k)%m]==0)
63         {
64             printf("%d",(x+k*k)%m);
65             h[(x+k*k)%m]=1;
66         }
67         else
68             printf("-");
69     }
70 
71     return 0;
72 }

优化： 

 1 #include<cstdio>
 2 const int M=1e5;
 3 
 4 bool IsPrime(int n)
 5 {
 6     if(n==0 || n==1)
 7         return false;
 8     for(int i=2; i*i<=n; i++)
 9         if(n%i==0)
10             return false;
11     return true;
12 }
13 
14 int main()
15 {
16 #ifdef ONLINE_JUDGE
17 #else
18     freopen("Test.txt", "r", stdin);
19 #endif // ONLINE_JUDGE
20 
21     int n,m,x,h[M]={0},prob[M]={0};
22     scanf("%d%d", &m,&n);
23     while(!IsPrime(m))
24         m++;
25     for(int i=0; i<n; i++)
26     {
27         scanf("%d", &x);
28         for(int j=prob[x%m]; j<=m; j++)
29         {
30             if(h[(x+j*j)%m]==0)
31             {
32                 h[(x+j*j)%m]=1;
33                 prob[x%m]=j+1;
34                 printf("%d%c", (x+j*j)%m,i==n-1?'\n':' ');
35                 x=-1;
36                 break;
37             }
38         }
39         if(x!=-1)
40             printf("-%c", i==n-1?'\n':' ');
41     }
42 
43     return 0;
44 }

 

posted @ 2019-05-13 21:04  林東雨  阅读(152)  评论(0)  编辑  收藏  举报