Project Euler Problem 9 (C++和Python代码实现和解析)

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本文探讨了在a+b+c=1000条件下寻找唯一满足a²+b²=c²的毕达哥拉斯三元组问题，通过数学分析和编程实现，最终求得三元组的乘积。

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Problem 9 : Special Pythagorean triplet

A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,

a² + b² = c²

For example, 3² + 4² = 9 + 16 = 25 = 5².

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

1. 第9个问题：特殊的毕达哥拉斯三元组

一个毕达哥拉斯三元组是一个包含三个自然数的集合(a, b, c)，且a < b < c和 a² + b² = c²。例如：3² + 4² = 9 + 16 = 25 = 5²。

唯一存在一个毕达哥拉斯三元组，且a+b+c=1000。求abc的乘积。

2. 求解分析

a+b+c = 1000, 且a < b < c和a² + b² = c²。我们可以得到 a < 1000/3，b < (1000-a)/2，就找到了两个循环的判断条件，然后根据c = 1000-a-b，就可以判断一下a² + b² 是否等于 c²，如果找到满足条件的唯一的毕达哥拉斯三元组，返回abc的乘积。

3. C++ 代码实现

在这里插入图片描述

函数checkPythagoreanTriplet()用来判断三元组a, b, c是否满足a² + b² = c²。

函数getProductOfPythagoreanTriplet()用来找到符合条件的a, b, c三元组。

C++ 代码

#include <iostream>
#include <cassert>

using namespace std;

// #define UNIT_TEST

class PE0009
{
public:
    bool checkPythagoreanTriplet(int a, int b, int c);
    int getProductOfPythagoreanTriplet(int sumOfabc);
};

bool PE0009::checkPythagoreanTriplet(int a, int b, int c)
{    
    if (a*a + b*b == c*c)
    {
        return true;
    }
    return false;
}

int PE0009::getProductOfPythagoreanTriplet(int sum)
{
    int c; // c = sum - a - b;    a < b < c
    int product = 0;
    
    for (int a=1; a<sum/3; a++)  // sum = a+b+c > 3a => a < sum/3
    {
        for (int b=a+1; b<(sum-a)/2; b++) // c = sum-a-b > b => b < (sum-a)/2
        {
            c = sum-a-b;
            if (true == checkPythagoreanTriplet(a, b, c))
            {
                product = a*b*c;
#ifdef UNIT_TEST
                cout << "The product abc(" << a << "," << b << "," << c << ")";
                cout << " is " << a*b*c << endl;
#endif
                return product;
            }
        }
    }
    return product;
}

int main()
{
    PE0009 pe0009;

    assert(true == pe0009.checkPythagoreanTriplet(3, 4, 5)); 
    
    cout << "There exists exactly one Pythagorean triplet for which " << endl;
    cout << "a + b + c = 1000, and the product abc is " ;
    cout << pe0009.getProductOfPythagoreanTriplet(1000) << "." << endl;
    
    return 0;     
}

4. Python 代码实现

Python实现方法跟C++类似的，也使用函数checkPythagoreanTriplet()来判断三元组a, b, c是否满足a² + b² = c²。函数getProductOfPythagoreanTriplet()用来找到满足条件的唯一三元组abc，并返回abc的乘积。

Python 代码

def checkPythagoreanTriplet(a, b, c):
    if c*c == a*a + b*b:
        return True
    else:
        return False

def getProductOfPythagoreanTriplet(p):
    """ 
    a < b < c and a+b+c = p
    p = a+b+c > 3a => a < p/3
    c = p-a-b > b  => b < (p-a)/2
    """
    product = 1
    for a in range (1, int(p/3)):
        for b in range(a+1, int((p-a)/2)):
            c = p-a-b
            if True == checkPythagoreanTriplet(a, b, c):
                product = a*b*c
                break
    return product
        
def main():
    assert True == checkPythagoreanTriplet(3, 4, 5)
    print("There exists exactly one Pythagorean triplet for which")
    print("a + b + c = 1000, and the product abc is %d."\
          % getProductOfPythagoreanTriplet(1000))

if  __name__ == '__main__':
    main()