通过抛N个骰子获得的值的乘积获得素数的可能性

给定一个表示骰子数量的整数N ，任务是找到出现在N个投掷骰子的顶面上的数字乘积是质数的概率。所有N个骰子必须同时投掷。

例子：

Input: N = 2 
Output: 6 / 36 
Explanation: 
On throwing N(=2) dices simultaneously, the possible outcomes on the top faces of N(=2) dices having product equal to a prime number are: {(1, 2), (1, 3), (1, 5), (2, 1), (3, 1), (5, 1)}. 
Therefore, the count of favourable outcomes = 6 and the count of the sample space is = 36 
Therefore, the required output is (6 / 36) 

Input: N = 3 
Output: 9 / 216

天真的方法：解决此问题的最简单方法是通过同时抛出N个骰子在N个骰子的顶面上生成所有可能的结果，并针对每个可能的结果检查顶面上的数字乘积是否为质数。如果发现为真，则增加计数器。最后，打印获得顶面上数字乘积作为质数的概率。

时间复杂度： O(6 ^N * N)
辅助空间： O(1)

高效的方法：为了优化上述方法，我们的想法是利用以下事实：只有(N – 1)个数字为1且其余数字为质数时， N个数的乘积才是质数。以下是观察结果：

If the product of N numbers is a prime number then the value of (N – 1) numbers must be 1 and the remaining number must be a prime number. 
Total count of prime numbers in the range [1, 6] is 3. 
Therefore, the total number of outcomes in which the product of N numbers on the top faces as a prime number = 3 * N.
P(E) = N(E) / N(S) 
P(E) = probability of getting the product of numbers on the top faces of N dices as a prime number. 
N(E) = total count of favourable outcomes = 3 * N 
N(S) = total number of events in the sample space = 6^N 
 

请按照以下步骤解决此问题：

• 初始化一个变量，例如N_E，以存储有利结果的计数。
• 初始化一个变量，例如N_S，以存储样本空间的数量。
• 更新N_E = 3 * N。
• 更新N_S = 6 ^N。
• 最后，打印(N_E / N_S)的值。

下面是上述方法的实现

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
 
// Function to find the value
// of power(X, N)
long long int power(long long int x,
                    long long int N)
{
    // Stores the value
    // of (X ^ N)
    long long int res = 1;
 
    // Calculate the value of
    // power(x, N)
    while (N > 0) {
         
       // If N is odd
       if(N & 1) {
            
           //Update res
           res = (res * x);
       }
        
       //Update x
       x = (x * x);
        
       //Update N
       N = N >> 1;
        
    }
    return res;
}
 
// Function to find the probability of
// obtaining a prime number as the
// product of N thrown dices
void probablityPrimeprod(long long int N)
{
    // Stores count of favorable outcomes
    long long int N_E = 3 * N;
     
    // Stores count of sample space
    long long int N_S = power(6, N);
     
    // Print the required probablity
    cout<

---

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
     
// Function to find the value
// of power(X, N)
static int power(int x, int N)
{
     
    // Stores the value
    // of (X ^ N)
    int res = 1;
 
    // Calculate the value of
    // power(x, N)
    while (N > 0)
    {
         
        // If N is odd
        if (N % 2 == 1)
        {
             
            // Update res
            res = (res * x);
        }
         
        // Update x
        x = (x * x);
         
        // Update N
        N = N >> 1;
    }
    return res;
}
 
// Function to find the probability of
// obtaining a prime number as the
// product of N thrown dices
static void probablityPrimeprod(int N)
{
     
    // Stores count of favorable outcomes
    int N_E = 3 * N;
     
    // Stores count of sample space
    int N_S = power(6, N);
     
    // Print the required probablity
    System.out.print(N_E + " / " + N_S);
}
 
// Driver code
public static void main(String[] args)
{
    int N = 2;
     
    probablityPrimeprod(N);
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach
 
# Function to find the value
# of power(X, N)
def power(x, N):
     
    # Stores the value
    # of (X ^ N)
    res = 1
 
    # Calculate the value of
    # power(x, N)
    while (N > 0):
 
        # If N is odd
        if (N % 2 == 1):
             
            # Update res
            res = (res * x)
 
        # Update x
        x = (x * x)
 
        # Update N
        N = N >> 1
 
    return res
 
# Function to find the probability of
# obtaining a prime number as the
# product of N thrown dices
def probablityPrimeprod(N):
     
    # Stores count of favorable outcomes
    N_E = 3 * N
 
    # Stores count of sample space
    N_S = power(6, N)
 
    # Prthe required probablity
    print(N_E, " / ", N_S)
 
# Driver code
if __name__ == '__main__':
     
    N = 2
 
    probablityPrimeprod(N)
 
# This code is contributed by 29AjayKumar

C#

// C# program to implement
// the above approach
using System;
class GFG{
     
// Function to find the
// value of power(X, N)
static int power(int x,
                 int N)
{   
  // Stores the value
  // of (X ^ N)
  int res = 1;
 
  // Calculate the value
  // of power(x, N)
  while (N > 0)
  {
    // If N is odd
    if (N % 2 == 1)
    {
      // Update res
      res = (res * x);
    }
 
    // Update x
    x = (x * x);
 
    // Update N
    N = N >> 1;
  }
  return res;
}
 
// Function to find the probability
// of obtaining a prime number as
// the product of N thrown dices
static void probablityPrimeprod(int N)
{   
  // Stores count of favorable
  // outcomes
  int N_E = 3 * N;
 
  // Stores count of sample
  // space
  int N_S = power(6, N);
 
  // Print the required
  // probablity
  Console.Write(N_E + " / " + N_S);
}
 
// Driver code
public static void Main(String[] args)
{
  int N = 2;
  probablityPrimeprod(N);
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// JavaScript program to implement the above approach
 
// Function to find the value
// of power(X, N)
function power(x, N)
{
     
    // Stores the value
    // of (X ^ N)
    let res = 1;
 
    // Calculate the value of
    // power(x, N)
    while (N > 0)
    {
         
        // If N is odd
        if (N % 2 == 1)
        {
             
            // Update res
            res = (res * x);
        }
         
        // Update x
        x = (x * x);
         
        // Update N
        N = N >> 1;
    }
    return res;
}
 
// Function to find the probability of
// obtaining a prime number as the
// product of N thrown dices
function probablityPrimeprod(N)
{
     
    // Stores count of favorable outcomes
    let N_E = 3 * N;
     
    // Stores count of sample space
    let N_S = power(6, N);
     
    // Print the required probablity
    document.write(N_E + " / " + N_S);
}
 
// Driver Code
    let N = 2;
    probablityPrimeprod(N);
     
    // This code is contributed by susmitakunndugoaldanga.
 

输出：

6 / 36

时间复杂度： O(log ₂ N)
辅助空间： O(1)