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Hashing (25)

[编程题]Hashing (25)

热度指数：7696 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 64M，其他语言128M
算法知识视频讲解

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.

输入描述:

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.

输出描述:

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.

示例1

输入

4 4
10 6 4 15

输出

0 1 4 -

算法知识视频讲解

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orgcheng

首先要理解英文题意
1. MSize是给的哈希表最大容量，如果它不是素数，需要自己找一个大于MSize的最小素数。

2. Quadratic probing翻译为二次探查法，数值插入哈希表遇到冲突时，需要通过二次探查的方式找到新的可插入位置，如果找不到，返回“-”。

对于给定的Key，求它所在位置的过程如下:

第一次：position = （key+ d1*d1） % MSzie; ( d1= 1)
第二次：position = （key+ d2*d2） % MSzie; ( d2= 2)
第三次：position = （key+ d3*d3） % MSzie; ( d3= 3)
第四次：position = （key+ d4*d4） % MSzie; ( d4= 4)
…… …… ……
第 i 次： position = （key+ di*di） % MSzie; ( di = i )
其中 1<= di < MSize，这里的MSize可能是修改后的值。

具体代码如下

import java.io.PrintStream;
import java.util.Scanner;

public class Main {
	public static Scanner in = new Scanner(System.in);
	public static PrintStream out = System.out;

	public static boolean isPrime(int n) {
		boolean flag = true;
		if (n < 2) {// 素数不小于2
			return false;
		} else {
			for (int i = 2; i <= Math.sqrt(n); i++) {
				if (n % i == 0) {// 若能被整除，则说明不是素数，返回false
					flag = false;
					break;// 跳出循环
				}
			}
		}
		return flag;
	}

	// 获取不小于n的最小素数
	public static int getMinPrime(int n) {
		while(true){
			if(isPrime(n))
				return n;
			++n;
		}
	}

	public static void main(String[] args) {
		int MSize = in.nextInt();
		MSize = getMinPrime(MSize);
		
		boolean[] visited = new boolean[MSize];
		int N = in.nextInt();
		
		int key = in.nextInt();
		int position =  key%MSize;
		out.print(position);
		visited[position] = true;
		
		for(int i=1;i<N;++i){
			key = in.nextInt();
			position =  key%MSize;
			
			if(visited[position]){
				int inc = 1;
				while(inc<MSize){
					if(!visited[(position+inc*inc)%MSize]){
						out.print(" "+(position+inc*inc)%MSize);
						visited[(position+inc*inc)%MSize] = true;
						break;
					}
					++inc;
				}
				if(inc >= MSize){
					out.print(" -");
				}
					
			}else{
				out.print(" "+position);
				visited[position] = true;
			}
		}
		out.println();
	}
}

编辑于 2015-08-18 21:37:14 回复(0)