zoukankan      html  css  js  c++  java
  • PAT-A 1078. Hashing

    1078. 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 "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 (<=104) 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 -
    

    哈希表的二次探测公式
    $$hash(key)=(key+i^2)%Tsize,i=0,1,2...$$

    程序代码:

    #include<iostream>
    #include<vector>
    #include<cmath>
    using namespace std;
    int NextPrime(int a);
    int main()
    {
    	int Tsize,N,tmp;
    	vector<int> num;
    	vector<int>position;
    	cin>>Tsize>>N;
    	Tsize = NextPrime(Tsize);
    	int p;
    	for(int i=0;i<N;i++)
    	{
    		cin>>tmp;
    		num.push_back(tmp);
    	}
    	for(int i=0;i<Tsize;i++)
    		position.push_back(-1);
    	
    	for(int i=0;i<N;i++)
    	{
    		tmp = num[i];
    		for(int j=0;;j++)
    		{
    			p = (tmp+(int)pow(j,2))%Tsize;
    			
    			if(position[p]==-1)
    			{
    				position[p]=1;
    				cout<<p;
    				break;
    			}
    			if(p==tmp%Tsize&&j!=0)
    			{
    				cout<<'-';
    				break;
    			}
    		}
    		if(i<N-1)
    			cout<<' ';
    	}
    }
    int NextPrime(int a)//求大于等于a的第一个素数
    {
    	int i;
    	if(a<2)
    		return 2;
    	for(i=2;i<=sqrt(a);i++)
    	{
    		if(a%i==0)
    		{
    			a++;
    			i=1;
    		}
    	}
    	return a;
    }
    
    
  • 相关阅读:
    htnl类名命规范
    JAVA集合中泛型的原理本质简介
    java 泛型中的上界(extend)和下界(super)
    elasticsearch深度分页问题
    AOP组合使用切面和自定义注解
    G1垃圾回收器基本知识及原理解析
    MyBatisplus源码解析
    生产环境碰到系统CPU飙高和频繁GC,你要怎么排查?
    fullgc触发条件_记一次生产频繁出现 Full GC 的 GC日志图文详解
    JavaThreadContextLoader(线程上线文类加载器)总结
  • 原文地址:https://www.cnblogs.com/zhengkang/p/5770587.html
Copyright © 2011-2022 走看看