📜  使用Pascal三角形计算nCr

📅  最后修改于: 2021-06-26 18:59:38             🧑  作者: Mango

Pascal三角形的一个有用的应用是组合的计算。寻找n C r的公式是n! / r! *(n – r)!这也是Pascal三角形单元格的公式。
帕斯卡的三角形:

Input: n = 5, r = 3
Output: 10
Explanation:
n! / r! * (n - r)! = 5! / 3! * (2!) = 120 / 12 = 10

Input: n = 7, r = 2
Output: 21
Explanation:
n! / r! * (n - r)! = 7! / 5! * (2!) = 42 / 2 = 21

的方法:我们的想法是帕斯卡三角存储在随后的矩阵的值n C R将是单元格的值在n行和r
要创建帕斯卡三角形,请使用以下两个公式:

  1. n C 0 = 1 ,从一组n个元素中选择0个元素的方式为0
  2. n C r = n-1 C r-1 + n-1 C r ,从一组n个元素中选择r个元素的方法数量是从n-1个元素中选择r-1个元素的方法和选择方法的总和n-1个元素中的r个元素。

想法是使用子问题的值来计算较大值的答案。例如,要计算n C r ,请使用n-1 C r-1n-1 C r的值。因此,DP可用于预处理范围内的所有值。
算法:

  1. 创建一个大小为1000 * 1000的矩阵,分配基本案例的值,即运行一个从0到1000的循环,并分配matrix [i] [0] = 1, n C 0 = 1
  2. 从i = 1到1000运行一个嵌套循环(外循环),从j = 1到i + 1运行内部循环。
  3. 使用公式n C r = n为每个元素(i,j)分配矩阵[i] [j] =矩阵[i-1] [j-1] +矩阵[i-1] [j]的-1 C r-1 + n-1 C r
  4. 填充矩阵后,将n C r的值作为矩阵[n] [r]

执行:

C++
// C++ implementation of the approach
#include 
using namespace std;
 
// Initialize the matrix with 0
int l[1001][1001] = { 0 };
 
void initialize()
{
 
    // 0C0 = 1
    l[0][0] = 1;
    for (int i = 1; i < 1001; i++) {
        // Set every nCr = 1 where r = 0
        l[i][0] = 1;
        for (int j = 1; j < i + 1; j++) {
 
            // Value for the current cell of Pascal's triangle
            l[i][j] = (l[i - 1][j - 1] + l[i - 1][j]);
        }
    }
}
 
// Function to return the value of nCr
int nCr(int n, int r)
{
    // Return nCr
    return l[n][r];
}
 
// Driver code
int main()
{
    // Build the Pascal's triangle
    initialize();
    int n = 8;
    int r = 3;
    cout << nCr(n, r);
}
 
// This code is contributed by ihritik


Java
// Java implementation of the approach
 
class GFG {
    // Initialize the matrix with 0
    static int l[][] = new int[1001][1001];
 
    static void initialize()
    {
 
        // 0C0 = 1
        l[0][0] = 1;
        for (int i = 1; i < 1001; i++) {
            // Set every nCr = 1 where r = 0
            l[i][0] = 1;
            for (int j = 1; j < i + 1; j++) {
 
                // Value for the current cell of Pascal's triangle
                l[i][j] = (l[i - 1][j - 1] + l[i - 1][j]);
            }
        }
    }
 
    // Function to return the value of nCr
    static int nCr(int n, int r)
    {
        // Return nCr
        return l[n][r];
    }
    // Driver code
    public static void main(String[] args)
    {
        // Build the Pascal's triangle
        initialize();
        int n = 8;
        int r = 3;
        System.out.println(nCr(n, r));
    }
}
 
// This code is contributed by ihritik


Python3
# Python3 implementation of the approach
 
# Initialize the matrix with 0
l = [[0 for i in range(1001)] for j in range(1001)]
 
def initialize():
     
    # 0C0 = 1
    l[0][0] = 1
    for i in range(1, 1001):
         
        # Set every nCr = 1 where r = 0
        l[i][0] = 1
        for j in range(1, i + 1):
             
            # Value for the current cell of Pascal's triangle
            l[i][j] = (l[i - 1][j - 1] + l[i - 1][j])
 
# Function to return the value of nCr
def nCr(n, r):
    # Return nCr
    return l[n][r]
 
# Driver code
# Build the Pascal's triangle
initialize()
n = 8
r = 3
print(nCr(n, r))


C#
// C# implementation of the approach
 
using System;
class GFG {
    // Initialize the matrix with 0
    static int[, ] l = new int[1001, 1001];
 
    static void initialize()
    {
 
        // 0C0 = 1
        l[0, 0] = 1;
        for (int i = 1; i < 1001; i++) {
            // Set every nCr = 1 where r = 0
            l[i, 0] = 1;
            for (int j = 1; j < i + 1; j++) {
 
                // Value for the current cell of Pascal's triangle
                l[i, j] = (l[i - 1, j - 1] + l[i - 1, j]);
            }
        }
    }
 
    // Function to return the value of nCr
    static int nCr(int n, int r)
    {
        // Return nCr
        return l[n, r];
    }
    // Driver code
    public static void Main()
    {
        // Build the Pascal's triangle
        initialize();
        int n = 8;
        int r = 3;
        Console.WriteLine(nCr(n, r));
    }
}
 
// This code is contributed by ihritik


输出:
56

复杂度分析:

  • 时间复杂度: O(1)。
    所有对的值都是预先计算的,因此回答查询的时间为O(1),尽管预计算会花费一些时间,但理论上预计算会花费固定的时间。
  • 空间复杂度: O(1)。
    需要恒定的空间。

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