不使用乘法的数字的阶乘

给定一个正数N ，任务是在不使用乘法运算符的情况下计算N的阶乘。

例子：

Input: N = 5
Output:
120
Explanation:
5*4*3*2*1=120

Input: N = 7
Output:
5040

观察：

A*B=A+A+A+A...B times.

This observation can be used as follows:
5!=5*4*3*2*1
=(5+5+5+5)*3*2*1
=(20+20+20)*2*1
=60+60
=120

方法：这个问题可以使用嵌套循环的概念来解决。可以使用另一个循环手动计算答案，而不是使用乘法运算符。请按照以下步骤解决问题：

1. 将变量ans初始化为N 。
2. 使用变量i从N-1迭代到1 ，然后执行以下操作：
   • 将变量sum初始化为0 。
   • 使用变量j从0迭代到i-1，并将ans添加到sum
   • 将sum添加到ans 。
3. 打印答案。

下面是上述方法的实现

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to calculate factorial of the number
// without using multiplication operator
int factorialWithoutMul(int N)
{
    // variable to store the final factorial
    int ans = N;
 
    // Outer loop
    for (int i = N - 1; i > 0; i--) {
        int sum = 0;
 
        // Inner loop
        for (int j = 0; j < i; j++)
            sum += ans;
        ans = sum;
    }
    return ans;
}
 
// Driver code
int main()
{
    // Input
    int N = 5;
 
    // Function calling
    cout << factorialWithoutMul(N) << endl;
    return 0;
}

Java

// Java program for the above approach
 
import java.io.*;
 
class GFG {
    // Function to calculate factorial of the number
    // without using multiplication operator
    public static int factorialWithoutMul(int N)
    {
        // variable to store the final factorial
        int ans = N;
 
        // Outer loop
        for (int i = N - 1; i > 0; i--) {
            int sum = 0;
 
            // Inner loop
            for (int j = 0; j < i; j++)
                sum += ans;
            ans = sum;
        }
        return ans;
    }
    // Driver code
    public static void main(String[] args)
    {
        int N = 5;
 
        // Function calling
        System.out.println(factorialWithoutMul(N));
        // This code is contributed by Potta Lokesh
    }
}

Python3

# Python3 program for the above approach
 
# Function to calculate factorial of the number
# without using multiplication operator
def factorialWithoutMul(N):
     
    # Variable to store the final factorial
    ans = N
 
    # Outer loop
    i = N - 1
     
    while (i > 0):
        sum = 0
 
        # Inner loop
        for j in range(i):
            sum += ans
             
        ans = sum
        i -= 1
         
    return ans
 
# Driver code
if __name__ == '__main__':
     
    # Input
    N = 5
 
    # Function calling
    print(factorialWithoutMul(N))
     
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to calculate factorial of the number
// without using multiplication operator
static int factorialWithoutMul(int N)
{
     
    // Variable to store the final factorial
    int ans = N;
 
    // Outer loop
    for(int i = N - 1; i > 0; i--)
    {
        int sum = 0;
 
        // Inner loop
        for(int j = 0; j < i; j++)
            sum += ans;
             
        ans = sum;
    }
    return ans;
}
 
// Driver code
public static void Main()
{
     
    // Input
    int N = 5;
 
    // Function calling
    Console.Write(factorialWithoutMul(N));
}
}
         
// This code is contributed by SURENDRA_GANGWAR

Javascript

// JavaScript program for the above approach
 
// Function to calculate factorial of the number
// without using multiplication operator
function factorialWithoutMul(N)
{
    // variable to store the final factorial
    let ans = N;
 
    // Outer loop
    for (let i = N - 1; i > 0; i--) {
        let sum = 0;
 
        // Inner loop
        for (let j = 0; j < i; j++)
            sum += ans;
        ans = sum;
    }
    return ans;
}
 
// Driver code
 
    // Input
    let N = 5;
 
    // Function calling
    document.write(factorialWithoutMul(N));
 
  // This code is contributed by Potta Lokesh
    

输出

120

时间复杂度： O(N ² )
辅助空间： O(1)