找出f(n)/ f(r)* f(nr)的值

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📜 找出f(n)/ f(r)* f(nr)的值

📅 最后修改于: 2021-06-26 22:45:27 🧑 作者: Mango

给定两个正整数n和r，其中n> r> 1。任务是找到f(n)/(f(r)* f(nr))的值。 F(n)定义如下：

1^-1 *2^-2 *3^-3 *….. n^-n

例子：

Input: n = 5, r = 3
Output: 1/200000

Input: n = 3, r = 2
Output: 1/27

解决此问题的一种简单方法是分别计算f(n)，f(r)和f(nr)，然后根据给定的公式计算结果，但是这会花费大量时间。
解决此问题的更好方法是在r和nr之间找到更大的值，然后使用给定的特性f(n)= f(n-1)* n ^-n = f(n-1)/ n ⁿ函数，从分子和分母中消除f(r)和f(nr)中的较大者。之后，使用简单的循环和幂函数计算其余的值。
算法：

find max(r, n-r).
iterate from max(r, n-r) to n
result = ((result * i^-i / (i-max(r, n-r)) ^{-(i-max(r, n-r))} )

为什么编程需要懂一点英语

下面是上述方法的实现：

C++

// CPP to find the value of f(n)/f(r)*f(n-r)
#include 
using namespace std;
 
// Function to find value of given F(n)
int calcFunction(int n, int r)
{
    int finalDenominator = 1;
    int mx = max(r, n - r);
 
    // iterate over n
    for (int i = mx + 1; i <= n; i++) {
 
        // calculate result
        int denominator = (int)pow(i, i);
        int numerator = (int)pow(i - mx, i - mx);
        finalDenominator = (finalDenominator
                            * denominator)
                           / numerator;
    }
 
    // return the result
    return finalDenominator;
}
 
// Driver code
int main()
{
    int n = 6, r = 2;
    cout << "1/" << calcFunction(n, r) << endl;
 
    return 0;
}

Java

// Java program to find the value of f(n)/f(r)*f(n-r) 
 
class GFG {
// Function to find value of given F(n)
 
    static int calcFunction(int n, int r) {
        int finalDenominator = 1;
        int mx = Math.max(r, n - r);
 
        // iterate over n
        for (int i = mx + 1; i <= n; i++) {
 
            // calculate result
            int denominator = (int) Math.pow(i, i);
            int numerator = (int) Math.pow(i - mx, i - mx);
            finalDenominator = (finalDenominator
                    * denominator)
                    / numerator;
        }
 
        // return the result
        return finalDenominator;
    }
 
// Driver code
    public static void main(String[] args) {
        int n = 6, r = 2;
        System.out.println("1/" + calcFunction(n, r));
 
    }
}
// This code is contributed by RAJPUT-JI

Python 3

# Python3 to find the value of f(n)/f(r)*f(n-r)
 
# Function to find value of given F(n)
def calcFunction(n, r):
 
    finalDenominator = 1
    mx = max(r, n - r)
 
    # iterate over n
    for i in range(mx + 1, n + 1):
 
        # calculate result
        denominator = pow(i, i)
        numerator = pow(i - mx, i - mx)
        finalDenominator = (finalDenominator *
                            denominator) // numerator
 
    # return the result
    return finalDenominator
 
# Driver code
if __name__ == "__main__":
     
    n = 6
    r = 2
    print("1/", end = "")
    print(calcFunction(n, r))
 
# This code is contributed by ita_c

C#

// C# program to find the value of f(n)/f(r)*f(n-r)
using System;
 
public class GFG {
// Function to find value of given F(n)
 
    static int calcFunction(int n, int r) {
        int finalDenominator = 1;
        int mx = Math.Max(r, n - r);
 
        // iterate over n
        for (int i = mx + 1; i <= n; i++) {
 
            // calculate result
            int denominator = (int) Math.Pow(i, i);
            int numerator = (int) Math.Pow(i - mx, i - mx);
            finalDenominator = (finalDenominator
                    * denominator)
                    / numerator;
        }
 
        // return the result
        return finalDenominator;
    }
 
// Driver code
   public static void Main() {
        int n = 6, r = 2;
        Console.WriteLine("1/" + calcFunction(n, r));
 
    }
}
// This code is contributed by RAJPUT-JI

PHP

Javascript

输出：

1/36450000