乘积小于 K 的子阵列数量
原文:https://www . geesforgeks . org/number-subarrays-product-less-k/
给定一个正数数组，计算乘积小于给定数 k 的可能连续子数组的数量
示例:
Input : arr[] = [1, 2, 3, 4]
K = 10
Output : 7
The subarrays are {1}, {2}, {3}, {4}
{1, 2}, {1, 2, 3} and {2, 3}
Input : arr[] = [1, 9, 2, 8, 6, 4, 3]
K = 100
Output : 16
Input : arr[] = [10, 5, 2, 6]
K = 100
Output : 8
这个问题的一个天真的方法是生成阵列的所有子阵列，然后计算乘积小于 k 的阵列的数量。 下面是上述方法的实现:
C++
// CPP program to count subarrays having
// product less than k.
#include <iostream>
using namespace std;
int countsubarray(int array[], int n, int k)
{
int count = 0;
int i, j, mul;
for (i = 0; i < n; i++) {
// Counter for single element
if (array[i] < k)
count++;
mul = array[i];
for (j = i + 1; j < n; j++) {
// Multiple subarray
mul = mul * array[j];
// If this multiple is less
// than k, then increment
if (mul < k)
count++;
else
break;
}
}
return count;
}
// Driver Code
int main()
{
int array[] = { 1, 2, 3, 4 };
int k = 10;
int size = sizeof(array) / sizeof(array[0]);
int count = countsubarray(array, size, k);
cout << count << "\n";
}
// This code is contributed by 'Dev Agarwal'.
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to count subarrays
// having product less than k.
class GFG {
static int countsubarray(int array[], int n, int k)
{
int count = 0;
int i, j, mul;
for (i = 0; i < n; i++) {
// Counter for single element
if (array[i] < k)
count++;
mul = array[i];
for (j = i + 1; j < n; j++) {
// Multiple subarray
mul = mul * array[j];
// If this multiple is less
// than k, then increment
if (mul < k)
count++;
else
break;
}
}
return count;
}
// Driver Code
public static void main(String args[])
{
int array[] = { 1, 2, 3, 4 };
int k = 10;
int size = array.length;
int count = countsubarray(array, size, k);
System.out.print(count);
}
}
// This code is contributed by Sam007
Python 3
# Python3 program to count subarrays
# having product less than k.
def countsubarray(array, n, k):
count = 0
for i in range(0, n):
# Counter for single element
if array[i] < k:
count += 1
mul = array[i]
for j in range(i + 1, n):
# Multiple subarray
mul = mul * array[j]
# If this multiple is less
# than k, then increment
if mul < k:
count += 1
else:
break
return count
# Driver Code
array = [1, 2, 3, 4]
k = 10
size = len(array)
count = countsubarray(array, size, k)
print(count, end=" ")
# This code is contributed by Shreyanshi Arun.
C
// C# program to count subarrays having
// product less than k.
using System;
public class GFG {
static int countsubarray(int[] array, int n, int k)
{
int count = 0;
int i, j, mul;
for (i = 0; i < n; i++) {
// Counter for single element
if (array[i] < k)
count++;
mul = array[i];
for (j = i + 1; j < n; j++) {
// Multiple subarray
mul = mul * array[j];
// If this multiple is less
// than k, then increment
if (mul < k)
count++;
else
break;
}
}
return count;
}
// Driver Code
static public void Main()
{
int[] array = { 1, 2, 3, 4 };
int k = 10;
int size = array.Length;
int count = countsubarray(array, size, k);
Console.WriteLine(count);
}
}
// This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to count subarrays
// having product less than k.
// function that returns count
function countsubarray($array, $n, $k)
{
$count = 0;
for ($i = 0; $i < $n; $i++)
{
// Counter for single element
if ($array[$i] < $k)
$count++;
$mul = $array[$i];
for ($j = $i + 1; $j < $n; $j++)
{
// Multiple subarray
$mul = $mul * $array[$j];
// If this multiple is less
// than k, then increment
if ($mul < $k)
$count++;
else
break;
}
}
return $count;
}
// Driver Code
$array = array(1, 2, 3, 4);
$k = 10;
$size = sizeof($array);
$count = countsubarray($array, $size, $k);
echo($count . "\n");
// This code is contributed by Ajit.
?>
java 描述语言
<script>
// javascript program to count subarrays
// having product less than k.
function countsubarray(array , n , k)
{
var count = 0;
var i, j, mul;
for (i = 0; i < n; i++)
{
// Counter for single element
if (array[i] < k)
count++;
mul = array[i];
for (j = i + 1; j < n; j++)
{
// Multiple subarray
mul = mul * array[j];
// If this multiple is less
// than k, then increment
if (mul < k)
count++;
else
break;
}
}
return count;
}
// Driver Code
var array = [ 1, 2, 3, 4 ];
var k = 10;
var size = array.length;
var count = countsubarray(array, size, k);
document.write(count);
// This code is contributed by todaysgaurav
</script>
Output
7
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