zoukankan      html  css  js  c++  java
  • 11-散列2 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%TSize( 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  (10​^4​​)and N(MSize) 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 -
    #include<cstdio>
    const int maxn = 10010;
    
    int hashTable[maxn] = {0};
    
    bool isPrime(int n)
    {
        if (n <= 1) return false;
        for (int i = 2; i * i <= n; i++)
        {
            if (n % i == 0)
            {
                return false;
            }
        }
        return true;
    }
    
    int main()
    {
        int n,m;
        scanf("%d%d", &m, &n);
        
        while(!isPrime(m))
        {
            m++;
        }
        
        int pos, temp;
        for (int i = 0 ; i < n; i++)
        {
            scanf("%d",&temp);
            pos = temp % m;
            if (!hashTable[pos])
            {
                hashTable[pos] = 1;
                printf("%d", pos);
            }
            else
            {
                int step = 1;
                for (; step <= m; step++)
                {
                    pos = (temp + step * step) % m; 
                    if (!hashTable[pos])
                    {
                        hashTable[pos] = 1;
                        printf("%d", pos);
                        break;
                    }
                }
                if (step >= m)
                {
                    printf("-");
                }
            }
            
            if (i < n - 1)
            {
                printf(" ");
            }
        }
        
        return 0;
    }
  • 相关阅读:
    python—Celery异步分布式
    python 安装虚拟环境步骤
    同源策略:
    git 的用法和命令
    一分钟搞定AlloyTouch图片轮播
    JavaScript学习总结(一)——延迟对象、跨域、模板引擎、弹出层、AJAX示例
    CSS3与页面布局学习总结(八)——浏览器兼容与前端性能优化
    移动前端UI选择
    你必须收藏的Github技巧
    HTML5 学习笔记(二)——HTML5新增属性与表单元素
  • 原文地址:https://www.cnblogs.com/wanghao-boke/p/11962862.html
Copyright © 2011-2022 走看看