Project Euler Problem 21 (C++和Python)

最新推荐文章于 2019-02-25 14:09:09 发布

原创最新推荐文章于 2019-02-25 14:09:09 发布 · 261 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#C++ #Project Euler #欧拉项目 #Python

C++ 同时被 3 个专栏收录

173 篇文章

订阅专栏

Python

100 篇文章

订阅专栏

Project Euler

87 篇文章

订阅专栏

本文介绍了一种数学问题，即寻找亲和数对，并提供了解决该问题的C++和Python代码实现。亲和数对是指两个数，其中一个数的所有真因子之和等于另一个数，反之亦然。文章详细解释了如何计算一个数的所有真因子之和，并通过代码展示了如何找出10000以下的所有亲和数对。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem 21 : Amicable numbers

Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.

For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.

Evaluate the sum of all the amicable numbers under 10000.

C++ source code

#include <iostream>
#include <cmath>     // sqrt()
#include <cassert>   // assert()

using namespace std;

class PE0021
{
private:
    int getSumOfProperDivisors(int number);

public:
    bool checkAmicableNumbers(int a, int b);
    int  getSumOfAmicableNumbers(int max);
};

int PE0021::getSumOfProperDivisors(int number)
{
    int sum = 1;

    for(int i=2; i<(int)sqrt((double)number); i++)
    {
        if (0 == number%i)
        {
            sum += i + number/i;
        }
        if (i*i == number)
        {
            sum -= i;
        }
    }
    return sum;
}

bool PE0021::checkAmicableNumbers(int a, int b)
{
    if (a == getSumOfProperDivisors(b) && 
        b == getSumOfProperDivisors(a))
    {
        return true;
    }
    else
    {
        return false;
    }
}

int PE0021::getSumOfAmicableNumbers(int max)
{
    int b;
    int sum = 0;

    for(int a=1; a<max; a++)
    {
        b = getSumOfProperDivisors(a);

        if (b != a && true == checkAmicableNumbers(a, b))
        {
            sum += a;
        }
    }
    return sum;
}

int main()
{
    PE0021 pe0021;

    assert(true == pe0021.checkAmicableNumbers(220, 284));

    cout << "The sum of all the amicable numbers under 10000 is ";
    cout << pe0021.getSumOfAmicableNumbers(10000) << endl; 
    
    return 0;
}

Python source code

import math
    
def d(n):
    sum  = 1
    root = math.sqrt(n)
    for i in range(2, int(root)+1):
        if n % i == 0: 
            sum += i + n/i
    if root == int(root): 
        sum -= root    #correct sum if n is a perfect square
    return sum

def checkAmicableNumbers(a, b):
    if (a == d(b) and b == d(a)):
        return True
    else:
        return False

def getSumOfAmicableNumbers(max):
    sum = 0
    for a in range(1, max):
        b = d(a)
        if b != a and True == checkAmicableNumbers(a, b):
            sum += a
    return sum

def main():
    assert True == checkAmicableNumbers(220, 284)
    print("The sum of all the amicable numbers under 10000 is",
          getSumOfAmicableNumbers(10000)) 

if  __name__ == '__main__':
    main()