Project Euler Problem 6 (C++和Python代码实现和解析)

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最新推荐文章于 2024-04-05 11:34:21 发布

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本文探讨了求解前一百个自然数的和的平方与平方之和的差的问题，通过C++和Python代码实现了两种不同的计算方法，即求平方之和与求和的平方，最后计算两者之间的差值。

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Problem 6 : Sum square difference

The sum of the squares of the first ten natural numbers is,
1² + 2² + … + 10² = 385

The square of the sum of the first ten natural numbers is,
(1 + 2 + … + 10)² = 55² = 3025

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

1. 第6个问题：和平方之差

前十个自然数的平方之和是：
1² + 2² + … + 10² = 385

前十个自然数之和的平方是：
(1 + 2 + … + 10)² = 55² = 3025

因此前十个自然数的和的平方与平方之和的差是： 3025 − 385 = 2640.

求前一百个自然数的和的平方与平方之和的差。

2. 求解分析

如果我们使用一个函数来计算前n个自然数的平方之和，另一个函数来计算前n个自然数的和的平方的话，就很容易得到两者的差了。

3. C++ 代码实现

在这里插入图片描述
类PE0006包含三个方法：getSumOfSquares(), getSquareOfSum()和getSumSquareDifference()。

函数getSumOfSquares()计算从1到n的平方之和。
函数getSquareOfSum()计算从1到n的和的平方。
函数getSumSquareDifference()计算从1到n的和的平方与平方之和的差。

C++ 代码

#include <iostream>
#include <cassert>

using namespace std;

class PE0006
{
private:
    int getSumOfSquares(int n);
    int getSquareOfSum(int n);

public:
    int getSumSquareDifference(int n);
};

int PE0006::getSumOfSquares(int n)
{
    // sumOfSquares = 1^2 + 2^2 + 3^2 + ... + n^2
    int sumOfSquares = 0;
    
    for(int i=1; i<=n; i++)
    {
        sumOfSquares += i*i;
    }
    
    return sumOfSquares;
}

int PE0006::getSquareOfSum(int n)
{
    // squareOfSum = (1 + 2 + 3 + ... + n)^2
    int sum = n*(n+1)/2;

    int squareOfSum = sum*sum;
        
    return squareOfSum ;
}

int PE0006::getSumSquareDifference(int n)
{
    return (getSquareOfSum(n) - getSumOfSquares(n));
}
    
int main()
{
    PE0006 pe0006;

    assert(2640 == pe0006.getSumSquareDifference(10));
    
    cout << "The difference between the sum of the squares of 1..100 and the ";
    cout << "square of the sum is " << pe0006.getSumSquareDifference(100) << endl;
    
    return 0;
}

4. Python 代码实现

Python实现方法跟C++一样，也包含三个函数。

Python 代码

def getSumOfSqures(n):
    """compute sum of squares from 1 to n^n """
    sumOfSquares = 0
    for i in range(1, n+1):
        sumOfSquares += i*i;
    return sumOfSquares

def getSqureOfSum(n):
    """compute square of sum from 1 to n """
    sum = (1+n)*n/2
    squareOfSum = sum*sum
    return squareOfSum

def getSumSquareDifference(n):
    return int(getSqureOfSum(n) - getSumOfSqures(n))

def main():
    assert 2640 == getSumSquareDifference(10)
    print("The difference between the sum of the squares of the first one ")
    print("hundred natural numbers and the square of the sum is ", end='')
    print(getSumSquareDifference(100), ".")
    
if  __name__ == '__main__':
    main()