Project Euler Problem 28 (C++和Python)

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探讨了在一个由数字构成的螺旋矩阵中，对角线上所有数字的总和计算方法。通过C++和Python代码实现，展示了如何计算5x5和1001x1001螺旋矩阵的对角线数字之和。

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Problem 28 : Number spiral diagonals

Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows:

21 22 23 24 25
20 7 8 9 10
19 6 1 2 11
18 5 4 3 12
17 16 15 14 13

It can be verified that the sum of the numbers on the diagonals is 101.

What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral formed in the same way?

C++ source code

#include <iostream>
#include <cassert>

using namespace std;

class PE0028 
{
public:
    long int getSumOfNumbersOnSpiralDiagonals(int N);
};

long int PE0028::getSumOfNumbersOnSpiralDiagonals(int N)
{
    // the law of the numbers:
    //
    // top right coner:   (2n+1)^2
    //
    // top left corner:   (2n+1)^2-(2n+1)+1
    // 
    // lower left corner: (2n+1)^2-4*n
    //
    // lower right corner:(2n+1)^2-6*n
    //
    // 2n+1<=N           : n<=(N-1)/2

    int square;
    long int sum = 1;  //  first number should be 1

    for(int n=1; n<=(N-1)/2; n++)
    {
        square = (2*n+1)*(2*n+1);

        sum += square;                // top right coner 
        sum += square - (2*n+1) + 1;  // top left corner 
        sum += square - 4*n;          // lower left corner
        sum += square - 6*n;          // lower right corner
    }

    return sum;
}

int main()
{
    PE0028 pe0028;

    assert(101 == pe0028.getSumOfNumbersOnSpiralDiagonals(5));

    cout << "The sum of the numbers on the diagonals in a 1001 by 1001 spiral ";
    cout << "formed in the same way is " ;
    cout << pe0028.getSumOfNumbersOnSpiralDiagonals(1001) << endl;

    return 0;

Python source code

def getSumOfNumbersOnSpiralDiagonals(N):
    """
    the law of the numbers:   
    top right coner   : (2n+1)^2    
    top left corner   : (2n+1)^2-(2n+1)+1
    lower left corner : (2n+1)^2-4*n    
    lower right corner: (2n+1)^2-6*n        
    2n+1<=N           :  n<=(N-1)/2
    """
    sum = 1  # first number should be 1
    for n in range(1, int((N-1)/2.0)+1):
        square = (2*n+1)*(2*n+1)
        sum += square                # top right coner 
        sum += square - (2*n+1) + 1  # top left corner
        sum += square - 4*n;         # lower left corner
        sum += square - 6*n;         # lower right corner
    return sum

def main():
    assert 101 == getSumOfNumbersOnSpiralDiagonals(5)
    print("The sum of the numbers on the diagonals in a 1001 by 1001 spiral")
    print("formed in the same way is",getSumOfNumbersOnSpiralDiagonals(1001))

if  __name__ == '__main__':
    main()