Question

三角数是从1到n的n个自然数之和。查找给定正整数是否为三角形的最快方法是什么？

这是第1200个到第1300个三角形数字的剪切，你可以在这里轻松看到一个位模式（如果没有，尝试缩小）：

(720600, '10101111111011011000')
(721801, '10110000001110001001')
(723003, '10110000100000111011')
(724206, '10110000110011101110')
(725410, '10110001000110100010')
(726615, '10110001011001010111')
(727821, '10110001101100001101')
(729028, '10110001111111000100')
(730236, '10110010010001111100')
(731445, '10110010100100110101')
(732655, '10110010110111101111')
(733866, '10110011001010101010')
(735078, '10110011011101100110')
(736291, '10110011110000100011')
(737505, '10110100000011100001')
(738720, '10110100010110100000')
(739936, '10110100101001100000')
(741153, '10110100111100100001')
(742371, '10110101001111100011')
(743590, '10110101100010100110')
(744810, '10110101110101101010')
(746031, '10110110001000101111')
(747253, '10110110011011110101')
(748476, '10110110101110111100')
(749700, '10110111000010000100')
(750925, '10110111010101001101')
(752151, '10110111101000010111')
(753378, '10110111111011100010')
(754606, '10111000001110101110')
(755835, '10111000100001111011')
(757065, '10111000110101001001')
(758296, '10111001001000011000')
(759528, '10111001011011101000')
(760761, '10111001101110111001')
(761995, '10111010000010001011')
(763230, '10111010010101011110')
(764466, '10111010101000110010')
(765703, '10111010111100000111')
(766941, '10111011001111011101')
(768180, '10111011100010110100')
(769420, '10111011110110001100')
(770661, '10111100001001100101')
(771903, '10111100011100111111')
(773146, '10111100110000011010')
(774390, '10111101000011110110')
(775635, '10111101010111010011')
(776881, '10111101101010110001')
(778128, '10111101111110010000')
(779376, '10111110010001110000')
(780625, '10111110100101010001')
(781875, '10111110111000110011')
(783126, '10111111001100010110')
(784378, '10111111011111111010')
(785631, '10111111110011011111')
(786885, '11000000000111000101')
(788140, '11000000011010101100')
(789396, '11000000101110010100')
(790653, '11000001000001111101')
(791911, '11000001010101100111')
(793170, '11000001101001010010')
(794430, '11000001111100111110')
(795691, '11000010010000101011')
(796953, '11000010100100011001')
(798216, '11000010111000001000')
(799480, '11000011001011111000')
(800745, '11000011011111101001')
(802011, '11000011110011011011')
(803278, '11000100000111001110')
(804546, '11000100011011000010')
(805815, '11000100101110110111')
(807085, '11000101000010101101')
(808356, '11000101010110100100')
(809628, '11000101101010011100')
(810901, '11000101111110010101')
(812175, '11000110010010001111')
(813450, '11000110100110001010')
(814726, '11000110111010000110')
(816003, '11000111001110000011')
(817281, '11000111100010000001')
(818560, '11000111110110000000')
(819840, '11001000001010000000')
(821121, '11001000011110000001')
(822403, '11001000110010000011')
(823686, '11001001000110000110')
(824970, '11001001011010001010')
(826255, '11001001101110001111')
(827541, '11001010000010010101')
(828828, '11001010010110011100')
(830116, '11001010101010100100')
(831405, '11001010111110101101')
(832695, '11001011010010110111')
(833986, '11001011100111000010')
(835278, '11001011111011001110')
(836571, '11001100001111011011')
(837865, '11001100100011101001')
(839160, '11001100110111111000')
(840456, '11001101001100001000')
(841753, '11001101100000011001')
(843051, '11001101110100101011')
(844350, '11001110001000111110')

例如，您是否也可以看到旋转的正态分布曲线，由807085和831405之间的零表示？这种模式经常重演.--＆gt;

Answer 1

如果n是m三角形数字，那么n = m*(m+1)/2。使用二次公式求解m：

m = (sqrt(8n+1) - 1) / 2

所以n是三角形的，当且仅当8n+1是完美的正方形时。要快速确定某个数字是否为完美正方形，请参阅此问题：Fastest way to determine if an integer’s square root is an integer。

注意，如果8n + 1是一个完美的正方形，那么上面公式中的分子将始终是偶数，所以不需要检查它是否可以被2整除。

Answer 2

如果8x + 1是正方形，则整数x是三角形。

Answer 3

家庭作业？

从1到N的数字总和

1 + 2 + 3 + 4 + ... n-1 + n

如果你添加第一个加上最后一个，然后第二个加上第二个加上最后一个，那么......

=（1 + n）+（2 + n-1）+（3 + n-2）+（4 + n-3）+ ...（n / 2 + n / 2 + 1）

=（n = 1）+（n + 1）+（n + 1）+ ...（n + 1）; .... n / 2次

= n（n + 1）/ 2

这应该让你开始......

Answer 4

我不知道这是否是最快的，但这里有一些数学可以让你朝着正确的方向......

S = n (n + 1) / 2
2*S = n^2 + n
n^2 + n - 2*S = 0

你现在有一个二次方程式。

解决n。

如果n没有小数位，你就可以了。

Answer 5

   int tri=(8*number)+1;// you can remove this for checking the perfect square and remaining all same for finding perfect square
   double trinum=Math.sqrt(tri);
int triround=(int) Math.ceil(trinum);
if((triround*triround)==tri)
     {
     System.out.println("Triangular");
     }
    else
    {
    System.out.println("Not Triangular");
    }

此处ceil始终会查看下一个最高编号，因此这将是进行验证的绝佳机会。

Answer 6

接受的答案将带您到另一个步骤，检查数字是否是完美的正方形。为什么不简单地遵循？找到一个完美的广场需要同样的努力。

public final static boolean isTriangularNumber(final long x)
{
    if (x < 0)
        return false;

    final long n = (long) Math.sqrt(2 * x);
    return n * (n + 1) / 2 == x;
}

Answer 7

我们只需要检查8 *（你要检查的整数）+1是否是一个完美的正方形！

public Boolean isSquare(int x)
{
    return(Math.sqrt(x)==(int)Math.sqrt(x));    // x will be a perfect square if and only if it's square root is an Integer.
}

}
public Boolean isTriangular(int z)
{
    return(isSquare(8*z+1));
}

Answer 8

要查找数字三角数是否使用以下公式：

const pagesInBook = (num) => Number.isInteger((Math.sqrt(8 * num+1) -1 )/2)

Answer 9

这是一个非常（最？）高效的 Python 变体（在 3.8 和 3.9 中）。

def isTriangular(n: int) -> bool:
    """
    when 8.0*n+1.0 is a perfect square it is triangular
    """
    x = 8.0*n+1.0
    return x % x**0.5 == 0.0

定义数字是否为三角形数字的最快方法

9 个答案: