由前 N 个自然数构成的集合的所有子集的乘积
原文:https://www . geesforgeks . org/由第一个 n 个自然数组成的集合的所有子集的乘积/
给定一个数 N ，任务是从由前 N 个自然数组成的集合的所有可能子集中找到所有元素的乘积。 例:
输入: N = 2 输出: 4 可能的子集有{{1}、{2}、{1，2}}。 子集内元素的乘积= {1} * {2} * {1 * 2} = 4 输入: N = 3 输出: 1296 可能的子集有{{1}、{2}、{3}、{1，2}、{1，3}、{2，3}、{1，2，3}} 子集内元素的乘积= 1 * 2 * 3 * (1 * 2) * (1
天真方法:一个简单的解决方案是生成前 N 个自然数的所有子集。然后对于每个子集，计算它的乘积，最后返回每个子集的总乘积。 高效方法:
- 可以观察到，原始数组的每个元素在所有子集中出现 2(N–1)次。
- 因此最终答案中任何元素arrIT3】的贡献都将是
i * 2(N – 1)
- 所以，所有子集的立方之和将是
12N-1 * 22N-1 * 32N-1......N2N-1
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the product of all elements
// in all subsets in natural numbers from 1 to N
int product(int N)
{
int ans = 1;
int val = pow(2, N - 1);
for (int i = 1; i <= N; i++) {
ans *= pow(i, val);
}
return ans;
}
// Driver Code
int main()
{
int N = 2;
cout << product(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG {
// Function to find the product of all elements
// in all subsets in natural numbers from 1 to N
static int product(int N)
{
int ans = 1;
int val = (int)Math.pow(2, N - 1);
for (int i = 1; i <= N; i++) {
ans *= (int)Math.pow(i, val);
}
return ans;
}
// Driver Code
public static void main (String[] args)
{
int N = 2;
System.out.println(product(N));
}
}
// This code is contributed by AnkitRai01
Python 3
# Python3 implementation of the approach
# Function to find the product of all elements
# in all subsets in natural numbers from 1 to N
def product(N) :
ans = 1;
val = 2 **(N - 1);
for i in range(1, N + 1) :
ans *= (i**val);
return ans;
# Driver Code
if __name__ == "__main__" :
N = 2;
print(product(N));
# This code is contributed by AnkitRai01
C
// C# implementation of the approach
using System;
class GFG {
// Function to find the product of all elements
// in all subsets in natural numbers from 1 to N
static int product(int N)
{
int ans = 1;
int val = (int)Math.Pow(2, N - 1);
for (int i = 1; i <= N; i++) {
ans *= (int)Math.Pow(i, val);
}
return ans;
}
// Driver Code
public static void Main (string[] args)
{
int N = 2;
Console.WriteLine(product(N));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// javascript implementation of the approach
// Function to find the product of all elements
// in all subsets in natural numbers from 1 to N
function product( N)
{
let ans = 1;
let val = Math.pow(2, N - 1);
for (let i = 1; i <= N; i++)
{
ans *= Math.pow(i, val);
}
return ans;
}
// Driver Code
let N = 2;
document.write(product(N));
// This code is contributed by todaysgaurav
</script>
Output:
4
时间复杂度: O(N*logN)
辅助空间: O(1)
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