三种颜色三角形

Question

我正在尝试为此问题制作代码:

(来源:https://www.codewars.com/kata/insane-coloured-triangles/train/c)

彩色三角形由一排颜色创建,每一种颜色都是红色,绿色或蓝色.通过考虑前一行中的两种触摸颜色,生成连续的行,每行包含比最后一种颜色少的颜色.如果这些颜色相同,则在新行中使用相同的颜色.如果它们不同,则在新行中使用缺少的颜色.这将持续到最后一行,只生成一种颜色.

例如,不同的可能性是:

Colour here:            G G        B G        R G        B R 
Becomes colour here:     G          R          B          G 

With a bigger example:
   R R G B R G B B  
    R B R G B R B
     G G B R G G
      G R G B G
       B B R R
        B G R
         R B
          G

您将获得三角形的第一行作为字符串,并且您的工作是返回最终颜色,该颜色将作为字符串显示在底行中.在上面的例子的情况下,你会被给予'RRGBRGBB',你应该返回'G'.

约束:

1 <= length(row) <= 10 ** 5

输入字符串只包含大写字母' B', 'G' or 'R'.

测试用例的确切数量如下:

100 tests of 100 <= length(row) <= 1000 
100 tests of 1000 <= length(row) <= 10000 
100 tests of 10000 <= length(row) <= 100000

例子:

triangle('B') == 'B'
triangle('GB') == 'R'
triangle('RRR') == 'R'
triangle('RGBG') == 'B'
triangle('RBRGBRB') == 'G'
triangle('RBRGBRBGGRRRBGBBBGG') == 'G'

所以我已经制作了这个代码,它适用于所有的样本品味,但是当它出现时length of row > 25由于我的阶乘功能而失败,并且在某些情况下的长度达到了100,000,所以任何建议至少可以解决这个问题的任何数学公式解决问题或小提示.

顺便说一句,在我找到一种解决此站点问题的数学方法后,我已经制作了这段代码:

https://math.stackexchange.com/questions/2257158/three-color-triangle-challenge

long long factorial(long long num)     
{
    long long fact = num;

    if (fact == 0)
        fact = 1;

    while (num > 1)
    {
        num--;
        fact *= num;
    }
    return (fact);
}

long long combination(long long n, long long k, long long fact_n)
{
    long long fact_k = factorial(k);
    long long comb = ( fact_n / (fact_k * factorial(n - k)) );
    return (comb);
}

char triangle(const char *row)
{
    long long len = strlen(row);
    long long n = len - 1;
    long long k = 0;
    int sign = pow(-1,n);
    long long sum = 0;
    char *result = "RGB";
    int a;
    long long fact_n = factorial(n);

    while (k < len)                 //This while calculate Segma (?). 
    {
        if (row[k] == 'R')
            a = 0;
        else if (row[k] == 'G')
            a = 1;
        else if (row[k] == 'B')
            a = 2;

        if (a != 0)
            sum = sum + (combination(n,k,fact_n) * a); 
        k++;
    }

    sum = sign * (sum % 3);

    if (sum == -1 || sum == -2)
        sum += 3;

  return (result[sum]);
}

Answer 1

我将假设您提供的链接中的公式是正确的:

为了避免整数溢出,我们需要应用这些模数算术规则:

(a * b) mod c = ((a mod c) * (b mod c)) mod c

(a ± b) mod c = ((a mod c) ± (b mod c)) mod c

将它们应用于公式:

但是我们如何计算

......没有直接计算系数本身(可能导致溢出)？

由于3是素数,这可以用卢卡斯定理来完成:

...其中n_i, m_i是i的个位数n, m的基底-3.

使用整数除法可轻松转换为base-3:

// convert a number to base 3
// and returns the number of digits
unsigned conv_base_3(unsigned n, unsigned max, unsigned* out)
{
    unsigned i = 0;
    while (i < max && n > 0)
    {
        out[i] = n % 3;
        n /= 3;
        i++;
    }
    return i;
}

请注意,因为n_i, m_i总是在范围内[0, 2](因为它们是基数为3位),C(n_i, m_i)所以很容易计算:

// calculate the binomial coefficient for n < 3
unsigned binom_max_2(unsigned n, unsigned k)
{
    if (n < k)
        return 0;
    switch (n)
    {
        case 0:
        case 1:
            return 1;
        case 2:
            return 1 + (k == 1);

        // shouldn't happen
        default:
            return 0;
    }
}

现在定理本身:

// Lucas's theorem for p = 3
unsigned lucas_3(
    unsigned len_n, const unsigned * dig_n,
    unsigned len_k, const unsigned * dig_k
)
{
    // use modulo product rule:
    // prod[i] % 3 = ((prod[i - 1] % 3) * value[i])      
    unsigned prod = 1;
    for (unsigned i = 0; i < len_n; i++) {
        unsigned n_i = dig_n[i];
        unsigned k_i = (i < len_k) ? dig_k[i] : 0;
        prod = (prod * binom_max_2(n_i, k_i)) % 3;
    }
    return prod % 3;
}

字符转换:

// convert from 012 to RGB
char int_2_char(int i)
{
    switch (i) {
        case 0: return 'R';
        case 1: return 'G';
        case 2: return 'B';

        // shouldn't happen
        default:
            return '\0';
    }
}

// convert from RGB to 012
unsigned char_2_int(char c)
{
    switch (c) {
        case 'R': return 0;
        case 'G': return 1;
        case 'B': return 2;

        // shouldn't happen
        default:
            return 3;
    }
}

最后,三角算法:

// the problem constraints state that n <= 10 ** 5
// max number of base-3 digits
#define MAX_N_LOG_3 11

// main algorithm function
char triangle(const char * input)
{
    unsigned sum = 0;
    const int n = strlen(input);

    // calculate digits of n - 1
    unsigned dig_n[MAX_N_LOG_3];
    unsigned len_n = conv_base_3(n - 1, MAX_N_LOG_3, dig_n);

    for (unsigned km1 = 0; km1 < n; km1++)
    {
        // calculate digits of k - 1
        unsigned dig_k[MAX_N_LOG_3];
        unsigned len_k = conv_base_3(km1, MAX_N_LOG_3, dig_k);

        // calculate C(n - 1, k - 1) mod 3
        unsigned Cnk_mod3 = lucas_3(len_n, dig_n, len_k, dig_k);

        // add using the modulo rule
        sum = (sum + Cnk_mod3 * char_2_int(input[km1])) % 3;
    }

    // value of (-1) ** (n - 1)
    // (no need for pow; just need to know if n is odd or even)
    int sign = (n % 2) * 2 - 1;

    // for negative numbers, must resolve the difference
    // between C's % operator and mathematical mod
    int sum_mod3 = (3 + (sign * (int)(sum % 3)) % 3;
    return int_2_char(sum_mod3);
}

上面的代码通过了所有测试; 请注意,它是为了清晰而不是表现而写的.

那么为什么这段代码能够在分配的时间内通过所有测试,而简单的基于表的方法呢？由于它的时间复杂性:

• 基于表格的方法处理三角形的所有级别,即O(n^2)(参见三角数).

• 当然,使用Lucas的算法,只需要处理顶层; 然而算法本身是O(log n),因为它遍历每个数字n(无论基数).总体复杂性O(n log n)仍然是一个显着的改进.