如何计数递归的次数?(计算X^n的乘法数）

How to count number of times recursion is called? (Computing number of multiplications for x^n)

本文关键字：递归 何计数 计算      更新时间：2023-10-16

我在弄清楚如何增加计数(在我的代码中(时遇到了一些困难，以至于power2的计数器等同于power1。我的程序以不同方式比较计算X^n的效率。我查找了有关如何计算递归呼唤次数的解决方案，但是无论我实现哪种方法，都会获得不正确的输出。任何帮助或准则将不胜感激！

这是我到目前为止我代码的一部分(Power1具有正确的计数器(：

template <class T>
T power1(T x, unsigned int n, unsigned int& mults)
{
mults = 0;
if (n == 0)
    return 1;
else if (n == 1)
    return x;
else
{
    T total = 1;
    for (int i = 0; i < n; ++i)
    {
        total = total * x;
        ++mults;
    }
    mults -= 1;
    return total;
}
}

template <class T>
T power2(T x, unsigned int n, unsigned int& mults)
{
++mults;
if (n == 0)
{
    mults = 0;
    return 1;
}
else if (n == 1)
    return x;
else
{
    if (n > 1)
    {
        return (x * power2(x, n - 1, mults));
    }
}
return x;
}

这是我输出的一部分：

Test for integer base:
2^0 = 1: mults1 = 0, mults2 = 0
2^1 = 2: mults1 = 0, mults2 = 1
2^2 = 4: mults1 = 1, mults2 = 3
2^3 = 8: mults1 = 2, mults2 = 6
2^4 = 16: mults1 = 3, mults2 = 10
2^5 = 32: mults1 = 4, mults2 = 15
2^6 = 64: mults1 = 5, mults2 = 21
2^7 = 128: mults1 = 6, mults2 = 28
2^8 = 256: mults1 = 7, mults2 = 36
2^9 = 512: mults1 = 8, mults2 = 45

您需要在呼叫之间重置MULTS2。

简单的解决方案是

power2(2,2,foo);
//reset before next call
foo=0;
power2(2,3,foo);

，也可以使函数自动重置。

template <class T>
T power2(T x, unsigned int n, unsigned int& mults, bool init = true)
{
    if (init)
        mults=0;
    if (n == 0)
    {
        mults = 0;
        return 1;
    }
    else if (n == 1)
        return x;
    else
    {
        if (n > 1)
        {
            ++mults;
            return (x * power2(x, n - 1, mults, false));
        }
    }
    return x;
}

我认为这里有两个问题：您在每个呼叫power2之前都不会重置多个问题，而您在开始时就会增加。这给出了正确的结果：

#include <iostream>
using namespace std;
template <class T>
T power2(T x, unsigned int n, unsigned int& mults)
{
    if (n == 0)
    {
        mults = 0;
        return 1;
    }
    else if (n == 1)
        return x;
    else
    {
        if (n > 1)
        {
            return (x * power2(x, n - 1, ++mults));
        }
    }
}
int main() {
    int x = 2;
    int n = 4;
    unsigned int mults = 0;
    int res = power2(x, n, mults);
    std::cout << x << "^" << n << " = " << res << " - mults: " << mults << "n";
    x = 2;
    n = 5;
    mults = 0;
    res = power2(2, 5, mults);
    std::cout << x << "^" << n << " = " << res << " - mults: " << mults << "n";
    return 0;
}

输出：

2^4 = 16 - mults: 3
2^5 = 32 - mults: 4
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