在 C 中计算嵌套根

问题描述

我被要求仅使用递归来计算以下嵌套根表达式.

I was asked to calculate the following nested root expression using recursion only.

我编写了以下有效的代码，但它们允许我们仅使用一个函数和 1 个输入 n 用于此目的，而不是像我使用的 2 个.有人能帮我把这段代码转换成一个可以计算表达式的函数吗?除了 中的函数外，不能使用任何库.

I wrote the code below that works, but they allowed us to use only one function and 1 input n for the purpose and not 2 like I used.Can someone help me transform this code into one function that will calculate the expression? cant use any library except functions from <math.h>.

n=10 的输出:1.757932

double rec_sqrt_series(int n, int m) {
    if (n <= 0)
        return 0;
    if (m > n)
        return 0;
    return sqrt(m + rec_sqrt_series(n, m + 1));
}

double helper(int n) {
    return rec_sqrt_series(n, 1);
}

推荐答案

使用 n 的高位作为计数器:

Use the upper bits of n as a counter:

double rec_sqrt_series(int n)
{
    static const int R = 0x10000;
    return n/R < n%R ? sqrt(n/R+1 + rec_sqrt_series(n+R)) : 0;
}

当然，当初始 n 是 R 或更大时，它会出现故障.这是一个更复杂的版本，适用于 n 的任何正值.它有效:

Naturally, that malfunctions when the initial n is R or greater. Here is a more complicated version that works for any positive value of n. It works:

• 当 n 为负数时，它的工作方式与上述版本类似，使用高位计数.
• 当n 为正数时，如果小于R，则用-n 调用自身，对上述函数求值.否则，它会在 R-1 否定的情况下调用自己.这会评估函数，就像它是用 R-1 调用的一样.这会产生正确的结果，因为在仅仅几十次迭代后，该系列就停止以浮点格式变化——更深的数字的平方根变得如此稀薄，以至于没有任何影响.因此，该函数对于所有 n 的值都相同，超过一个小阈值.

• When n is negative, it works like the above version, using the upper bits to count.
• When n is positive, if it is less than R, it calls itself with -n to evaluate the function as above. Otherwise, it calls itself with R-1 negated. This evaluates the function as if it were called with R-1. This produces the correct result because the series stops changing in the floating-point format after just a few dozen iterations—the square roots of the deeper numbers get so diluted they have no effect. So the function has the same value for all n over a small threshold.

double rec_sqrt_series(int n)
{
    static const int R = 0x100;
    return
        0 < n ? n < R ? rec_sqrt_series(-n) : rec_sqrt_series(1-R)
              : n/R > n%R ? sqrt(-n/R+1 + rec_sqrt_series(n-R)) : 0;
}

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