以下是对 二项式系数 .
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- A. 二项式系数 C(n,k)可以定义为(1+X)^n展开式中X^k的系数。
- 二项式系数C(n,k)也给出了从n个对象中选择k个对象的方法的数量,而不考虑顺序;更正式地说,n-元素集的k-元素子集(或k-组合)的数量。
给定两个数字n和r,求 N C R 例如:
Input : n = 5, r = 2Output : 10The value of 5C2 is 10Input : n = 3, r = 1Output : 3
这个想法只是基于以下公式。
N C R =(n!)/(r!*(n-r)!)
C
#include <stdio.h> int factorial( int n) { int factorial = 1; for ( int i = 2; i <= n; i++) factorial = factorial * i; return factorial; } int nCr( int n, int r) { return factorial(n) / (factorial(r) * factorial(n - r)); } int main() { int n = 5, r = 3; printf ( "%d" , nCr(n, r)); return 0; } // This code was contributed by Omkar Prabhune |
C++
// CPP program To calculate The Value Of nCr #include <bits/stdc++.h> using namespace std; int fact( int n); int nCr( int n, int r) { return fact(n) / (fact(r) * fact(n - r)); } // Returns factorial of n int fact( int n) { int res = 1; for ( int i = 2; i <= n; i++) res = res * i; return res; } // Driver code int main() { int n = 5, r = 3; cout << nCr(n, r); return 0; } |
JAVA
// Java program To calculate // The Value Of nCr class GFG { static int nCr( int n, int r) { return fact(n) / (fact(r) * fact(n - r)); } // Returns factorial of n static int fact( int n) { int res = 1 ; for ( int i = 2 ; i <= n; i++) res = res * i; return res; } // Driver code public static void main(String[] args) { int n = 5 , r = 3 ; System.out.println(nCr(n, r)); } } // This code is Contributed by // Smitha Dinesh Semwal. |
Python 3
# Python 3 program To calculate # The Value Of nCr def nCr(n, r): return (fact(n) / (fact(r) * fact(n - r))) # Returns factorial of n def fact(n): res = 1 for i in range ( 2 , n + 1 ): res = res * i return res # Driver code n = 5 r = 3 print ( int (nCr(n, r))) # This code is contributed # by Smitha |
C#
// C# program To calculate // The Value Of nCr using System; class GFG { static int nCr( int n, int r) { return fact(n) / (fact(r) * fact(n - r)); } // Returns factorial of n static int fact( int n) { int res = 1; for ( int i = 2; i <= n; i++) res = res * i; return res; } // Driver code public static void Main() { int n = 5, r = 3; Console.Write(nCr(n, r)); } } // This code is Contributed by nitin mittal. |
PHP
<?php // PHP program To calculate // the Value Of nCr function nCr( $n , $r ) { return fact( $n ) / (fact( $r ) * fact( $n - $r )); } // Returns factorial of n function fact( $n ) { $res = 1; for ( $i = 2; $i <= $n ; $i ++) $res = $res * $i ; return $res ; } // Driver code $n = 5; $r = 3; echo nCr( $n , $r ); // This code is contributed by vt_m. ?> |
Javascript
<script> // Javascript program To calculate The Value Of nCr function nCr(n, r) { return fact(n) / (fact(r) * fact(n - r)); } // Returns factorial of n function fact(n) { var res = 1; for ( var i = 2; i <= n; i++) res = res * i; return res; } // Driver code var n = 5, r = 3; document.write(nCr(n, r)); </script> |
输出:
10
时间复杂性: O(n)
辅助空间: O(1)
更高效的解决方案: 动态规划|集9(二项式系数) 时空效率二项式系数 关于二项式系数的所有文章
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