给定n个不同元素的数组，计算子集的总数。
例子：
Input : {1, 2, 3}
Output : 8
Explanation
the array contain total 3 element.its subset
are {}, {1}, {2}, {3}, {1, 2}, {2, 3}, {3, 1}, {1, 2, 3}.
so the output is 8..我们知道大小为n的集合的子集数为2 n
这个公式如何运作?
对于每个元素，我们都有两个选择，我们要么选择它，要么不选择它。因此，总共我们有2 * 2 *…(n次)个选择，即2 n
替代解释是：
大小为0的子集数量= n C 0
大小为1的子集数量= n C 1
大小为2的子集数量= n C 2
………………..
子集总数= n C 0 + n C 1 + n C 2 +…。 + n C n = 2 n
有关详细信息，请参考二项式系数之和。
C++
// CPP program to count number of distinct
// subsets in an array of distinct numbers
#include
using namespace std;
// Returns 2 ^ n
int subsetCount(int arr[], int n)
{
return 1 << n;
}
/* Driver program to test above function */
int main()
{
int A[] = { 1, 2, 3 };
int n = sizeof(A) / sizeof(A[0]);
cout << subsetCount(A, n);
return 0;
} Java
// Java program to count number of distinct
// subsets in an array of distinct numbers
class GFG {
// Returns 2 ^ n
static int subsetCount(int arr[], int n)
{
return 1 << n;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int A[] = { 1, 2, 3 };
int n = A.length;
System.out.println(subsetCount(A, n));
}
}
// This code is contributed by Prerna Saini.Python3
# Python3 program to count number
# of distinct subsets in an
# array of distinct numbers
import math
# Returns 2 ^ n
def subsetCount(arr, n):
return 1 << n
# driver code
A = [ 1, 2, 3 ]
n = len(A)
print(subsetCount(A, n))
# This code is contributed by Gitanjali.C#
// C# program to count number of distinct
// subsets in an array of distinct numbers
using System;
class GFG {
// Returns 2 ^ n
static int subsetCount(int []arr, int n)
{
return 1 << n;
}
// Driver program
public static void Main()
{
int []A = { 1, 2, 3 };
int n = A.Length;
Console.WriteLine(subsetCount(A, n));
}
}
// This code is contributed by vt_m.PHP
Javascript
输出：
8