算法设计与分析: 1-3 最多约数问题

1-3 最多约数问题

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问题描述

正整数x的约数是能整除x的正整数。正整数x的约数个数记为$div(x)$。例如，1,2,5,10都是正整数10的约数，且div(10)=4。 对于给定的2个正整数a<=b，编程计算a与b之间约数个数最多的数。

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分析

设正整数x的质因子分解为 $x=p_1^{N_1}p_2^{N_2}...p_k^{N_k}$，则
$div(x)=(N_1+1)(N_2+1)...(N_k+1)$

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Java

import java.util.Scanner;

public class Main {

    //1<=a<=b
    private static int MAXP = 100005;
    private static int PCOUNT = 0;//when MAXP=100005, we have PCOUNT=9593, then the largest prime number is prime[9592]=100003
    private static int[] prime = new int[MAXP];
    private static boolean[] isPrime = new boolean[MAXP];

    public static void main(String[] args) {
//        primes();
        optimisedPrimes();
        Scanner input = new Scanner(System.in);

        while (true){
            System.out.println("Please input the value of a:");
            int a = input.nextInt();
            System.out.println("Please input the value of b:");
            int b = input.nextInt();

            int max = 0;
            int count = 0;
            int minX = 0;
            for(int i=a; i<=b; i++){
                count = div(i);
                if(count > max){
                    max = count;
                    minX = i;
                }
            }

            System.out.println("There are at most "+max+" divisors of the number between "+a+" and "+b);
            System.out.println("The smallest number which has the most divisors is "+minX);
            System.out.println("------------------------------");
        }

    }

    //get the number of divisors in number x
    private static int div(int x){
        int count;
        int sum=1;
        for(int i=0; i<PCOUNT && prime[i]<=x; i++){
            count = 1;
            if((x%prime[i]) == 0){
                do{
                    count++;
                    x /= prime[i];
                }while ((x%prime[i]) == 0);
                sum *= count;
            }
        }

        return sum;
    }

    //build primes array
    private static void primes(){
        int i,j;
        for(i=0; i<MAXP; i++){
            isPrime[i] = true;
        }

        for(i=2; i<MAXP; i++){
            if(isPrime[i]){
                j = i + i;
                while (j < MAXP){
                    isPrime[j] = false;
                    j += i;
                }
            }
        }

        for(i=2; i<MAXP; i++){
            if(isPrime[i]){
                prime[PCOUNT++] = i;
            }
        }

    }

    //build optimised primes array
    private static void optimisedPrimes()
    {
        int i,j;
        for(i=0; i<MAXP; i++){
            isPrime[i] = true;
        }

        for(i=2; i<Math.sqrt(MAXP); i++){
            if(isPrime[i]){
                j = i + i;
                while (j < MAXP){
                    isPrime[j] = false;
                    j += i;
                }
            }
        }

        for(i=2; i<MAXP; i++){
            if(isPrime[i]){
                prime[PCOUNT++] = i;
            }
        }
    }

}

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Sample input & output

Please input the value of a:
1
Please input the value of b:
36
There are at most 9 divisors of the number between 1 and 36
The smallest number which has the most divisors is 36
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Please input the value of a:
24
Please input the value of b:
25
There are at most 8 divisors of the number between 24 and 25
The smallest number which has the most divisors is 24
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Please input the value of a:
1
Please input the value of b:
100005
There are at most 128 divisors of the number between 1 and 100005
The smallest number which has the most divisors is 83160
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Please input the value of a:

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Reference

王晓东《计算机算法设计与分析》（第3版）P7