Armstrong数字

题目：在3位的正整数中，比如abc，有一些可以满足a3 + b3 + c3 = abc的条件，也就是说，各个位数的立方和正好是该数的本身，这些数就叫做Armstrong数。试写一个程序求出所有的3位Armstrong数

#include <iostream>

using namespace std;

void printArmstrong()
{
	int leftValue;
	int midValue;
	int rightValue;
	for (int i = 100; i <= 999; i++)
	{
		leftValue = i / 100;
		midValue = (i / 10) % 10;
		rightValue = i % 10;

		if (leftValue * leftValue * leftValue + midValue * midValue * midValue + rightValue * rightValue * rightValue == i)
		{
			cout << "armstrong: " << i << endl;
		}
	}
}

void main()
{
	printArmstrong();
}

#include <iostream>

using namespace std;

void printArmstrong()
{
	for (int i = 1, i100 = 100; i <= 9; i++, i100 += 100)
	{
		int leftCube = i * i * i;
		for (int j = 0, j10 = 0; j <= 9; j++, j10 += 10)
		{
			int midCube = j * j * j;
			for (int k = 0, k0 = 0; k <= 9; k++, k0++)
			{
				int rightCube = k * k * k;
				if ((i100 + j10 + k0) == (leftCube + midCube + rightCube))
					cout << "armstrong: " << leftCube + midCube + rightCube << endl;
			}
		}
	}
}

void main()
{
	printArmstrong();
}

利用公式(p + 1)^3 = p^3 + 3p^2 + 3p + 1 和 (p + 1)^2 = p^2 + 2p + 1对上述代码进行优化：

#include <iostream>
#include <iterator>
#include <algorithm>

using namespace std;

int squareArray[10];
int cubeArray[10];

void printArmstrong()
{
	*squareArray = 0;
	*cubeArray = 0;
	for (int i = 1; i < 10; i++)
	{
		squareArray[i] = squareArray[i - 1] + 2 * (i - 1) + 1;
	}
	for (int i = 1; i < 10; i++)
	{
		cubeArray[i] = cubeArray[i - 1] + 3 * squareArray[i - 1] + 3 * (i - 1) + 1;
	}
	for (int i = 1, i100 = 100; i <= 9; i++, i100 += 100)
	{
		int leftCube = cubeArray[i];
		for (int j = 0, j10 = 0; j <= 9; j++, j10 += 10)
		{
			int midCube = cubeArray[j];
			for (int k = 0, k0 = 0; k <= 9; k++, k0++)
			{
				int rightCube = cubeArray[k];
				if ((i100 + j10 + k0) == (leftCube + midCube + rightCube))
					cout << "armstrong: " << leftCube + midCube + rightCube << endl;
			}
		}
	}
}

void main()
{
	printArmstrong();
}