一个阵列的所有子阵列的乘积
给定一个整数数组 arr 的大小 N ，任务是打印数组的所有子数组的产品。
示例:
输入: arr[] = {2，4 } T3】输出: 64 这里，子阵是【2】、【2，4】、【4】 积是所有子阵的 2，8，4 积= 64
输入: arr[] = {10，3，7} 输出: 27783000 这里，子阵是[10]，[10，3]，[10，3，7]，[3，7]，[7] 产品是所有子阵的 10，30，210，3，21，7 产品= 27783000
天真方法:一个简单的解决方案是生成所有子阵列并计算它们的乘积。
C++
// C++ program to find product
// of all subarray of an array
#include <bits/stdc++.h>
using namespace std;
// Function to find product of all subarrays
void product_subarrays(int arr[], int n)
{
// Variable to store the product
int product = 1;
// Compute the product while
// traversing for subarrays
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
for (int k = i; k <= j; k++)
product *= arr[k];
}
}
// Printing product of all subarray
cout << product << "\n";
}
// Driver code
int main()
{
int arr[] = { 10, 3, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
product_subarrays(arr, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find product
// of all subarray of an array
import java.util.*;
class GFG {
// Function to find product of all subarrays
static void product_subarrays(int arr[], int n)
{
// Variable to store the product
int product = 1;
// Compute the product while
// traversing for subarrays
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
for (int k = i; k <= j; k++)
product *= arr[k];
}
}
// Printing product of all subarray
System.out.print(product + "\n");
}
// Driver code
public static void main(String args[])
{
int arr[] = { 10, 3, 7 };
int n = arr.length;
// Function call
product_subarrays(arr, n);
}
}
// This code is contributed by shivanisinghss2110
Python 3
# Python3 program to find product
# of all subarray of an array
# Function to find product of all subarrays
def product_subarrays(arr, n):
# Variable to store the product
product = 1;
# Compute the product while
# traversing for subarrays
for i in range(0, n):
for j in range(i, n):
for k in range(i, j + 1):
product *= arr[k];
# Printing product of all subarray
print(product, "\n");
# Driver code
arr = [ 10, 3, 7 ];
n = len(arr);
# Function call
product_subarrays(arr, n);
# This code is contributed by Code_Mech
C
// C# program to find product
// of all subarray of an array
using System;
class GFG {
// Function to find product of all subarrays
static void product_subarrays(int[] arr, int n)
{
// Variable to store the product
int product = 1;
// Compute the product while
// traversing for subarrays
for (int i = 0; i < n; i++) {
for (int j = i; j < n; j++) {
for (int k = i; k <= j; k++)
product *= arr[k];
}
}
// Printing product of all subarray
Console.Write(product + "\n");
}
// Driver code
public static void Main(String[] args)
{
int[] arr = { 10, 3, 7 };
int n = arr.Length;
// Function call
product_subarrays(arr, n);
}
}
// This code is contributed by shivanisinghss2110
java 描述语言
<script>
// Javascript program to find product
// of all subarray of an array
// Function to find product of all subarrays
function product_subarrays(arr, n)
{
// Variable to store the product
let product = 1;
// Compute the product while
// traversing for subarrays
for (let i = 0; i < n; i++) {
for (let j = i; j < n; j++) {
for (let k = i; k <= j; k++)
product *= arr[k];
}
}
// Printing product of all subarray
document.write(product + "</br>");
}
let arr = [ 10, 3, 7 ];
let n = arr.length;
// Function call
product_subarrays(arr, n);
// This code is contributed by divyeshrabadiya07.
</script>
Output:
27783000
时间复杂度:O(n 3 )
辅助空间:0(1)
有效方法:一种有效方法是使用两个循环并在遍历子阵列时计算乘积。 以下是上述办法的实施情况:
C++
// C++ program to find product
// of all subarray of an array
#include <bits/stdc++.h>
using namespace std;
// Function to find product of all subarrays
void product_subarrays(long long int arr[], int n)
{
// Variable to store the product
long long int res = 1;
// Compute the product while
// traversing for subarrays
for (int i = 0; i < n; i++) {
long long int product = 1;
for (int j = i; j < n; j++) {
product = product * arr[j];
res *= product;
}
}
// Printing product of all subarray
cout << res << "\n";
}
// Driver code
int main()
{
long long int arr[] = { 10, 3, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
product_subarrays(arr, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find product
// of all subarray of an array
import java.util.*;
class GFG {
// Function to find product of all subarrays
static void product_subarrays(int arr[], int n)
{
// Variable to store the product
int res = 1;
// Compute the product while
// traversing for subarrays
for (int i = 0; i < n; i++) {
int product = 1;
for (int j = i; j < n; j++) {
product = product * arr[j];
res *= product;
}
}
// Printing product of all subarray
System.out.println(res + "\n");
}
// Driver code
public static void main(String args[])
{
int arr[] = { 10, 3, 7 };
int n = arr.length;
// Function call
product_subarrays(arr, n);
}
}
// This code is contributed by AbhiThakur
Python 3
# Python3 program to find product
# of all subarray of an array
# Function to find product of all subarrays
def product_subarrays(arr, n):
# Variable to store the product
res = 1;
# Compute the product while
# traversing for subarrays
for i in range(n):
product = 1
for j in range(i, n):
product *= arr[j];
res = res * product
# Printing product of all subarray
print(res);
# Driver code
if __name__ == '__main__':
arr = [ 10, 3, 7 ];
n = len(arr);
# Function call
product_subarrays(arr, n);
# This code is contributed by Princi Singh
C
// C# program to find product
// of all subarray of an array
using System;
class GFG {
// Function to find product of all subarrays
static void product_subarrays(int[] arr, int n)
{
// Variable to store the product
int res = 1;
// Compute the product while
// traversing for subarrays
for (int i = 0; i < n; i++) {
int product = 1;
for (int j = i; j < n; j++) {
product *= arr[j];
res = res * product;
}
}
// Printing product of all subarray
Console.WriteLine(res + "\n");
}
// Driver code
public static void Main(String[] args)
{
int[] arr = { 10, 3, 7 };
int n = arr.Length;
// Function call
product_subarrays(arr, n);
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript program to find product
// of all subarray of an array
// Function to find product of all subarrays
function product_subarrays(arr, n)
{
// Variable to store the product
var res = 1;
// Compute the product while
// traversing for subarrays
for (var i = 0; i < n; i++) {
var product = 1;
for (var j = i; j < n; j++) {
product = product * arr[j];
res *= product;
}
}
// Printing product of all subarray
document.write( res );
}
// Driver code
var arr = [10, 3, 7];
var n = arr.length;
// Function call
product_subarrays(arr, n);
</script>
Output:
27783000
时间复杂度:O(n 2 )
辅助空间:0(1)
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