给出一个由数组{0,1}中的元素组成的整数数组arr [] 。任务是打印将数组划分为子数组的方式的数量,以使每个子数组恰好包含一个1 。
例子:
Input: arr[] = {1, 0, 1, 0, 1}
Output: 4
Below are the possible ways:
- {1, 0}, {1, 0}, {1}
- {1}, {0, 1, 0}, {1}
- {1, 0}, {1}, {0, 1}
- {1}, {0, 1}, {0, 1}
Input: arr[] = {0, 0, 0}
Output: 0
方法:
- 当数组的所有元素均为0时,结果将为零。
- 否则,两个相邻的对象之间必须只有一个分隔。因此,答案等于值pos i + 1 -pos i (对于所有有效对)的乘积,其中pos i是第i 1个位置。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the number of ways
// the array can be divided into sub-arrays
// satisfying the given condition
int countWays(int arr[], int n)
{
int pos[n], p = 0, i;
// for loop for saving the positions of all 1s
for (i = 0; i < n; i++) {
if (arr[i] == 1) {
pos[p] = i + 1;
p++;
}
}
// If array contains only 0s
if (p == 0)
return 0;
int ways = 1;
for (i = 0; i < p - 1; i++) {
ways *= pos[i + 1] - pos[i];
}
// Return the total ways
return ways;
}
// Driver code
int main()
{
int arr[] = { 1, 0, 1, 0, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << countWays(arr, n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Function to return the number of ways
// the array can be divided into sub-arrays
// satisfying the given condition
static int countWays(int arr[], int n)
{
int pos[] = new int[n];
int p = 0, i;
// for loop for saving the
// positions of all 1s
for (i = 0; i < n; i++)
{
if (arr[i] == 1)
{
pos[p] = i + 1;
p++;
}
}
// If array contains only 0s
if (p == 0)
return 0;
int ways = 1;
for (i = 0; i < p - 1; i++)
{
ways *= pos[i + 1] - pos[i];
}
// Return the total ways
return ways;
}
// Driver code
public static void main(String args[])
{
int[] arr = { 1, 0, 1, 0, 1 };
int n = arr.length;
System.out.println(countWays(arr, n));
}
}
// This code is contributed
// by Akanksha Rai
Python3
# Python 3 implementation of the approach
# Function to return the number of ways
# the array can be divided into sub-arrays
# satisfying the given condition
def countWays(are, n):
pos = [0 for i in range(n)]
p = 0
# for loop for saving the positions
# of all 1s
for i in range(n):
if (arr[i] == 1):
pos[p] = i + 1
p += 1
# If array contains only 0s
if (p == 0):
return 0
ways = 1
for i in range(p - 1):
ways *= pos[i + 1] - pos[i]
# Return the total ways
return ways
# Driver code
if __name__ == '__main__':
arr = [1, 0, 1, 0, 1]
n = len(arr)
print(countWays(arr, n))
# This code is contributed by
# Surendra_Gangwar
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to return the number of ways
// the array can be divided into sub-arrays
// satisfying the given condition
static int countWays(int[] arr, int n)
{
int[] pos = new int[n];
int p = 0, i;
// for loop for saving the positions
// of all 1s
for (i = 0; i < n; i++)
{
if (arr[i] == 1)
{
pos[p] = i + 1;
p++;
}
}
// If array contains only 0s
if (p == 0)
return 0;
int ways = 1;
for (i = 0; i < p - 1; i++)
{
ways *= pos[i + 1] - pos[i];
}
// Return the total ways
return ways;
}
// Driver code
public static void Main()
{
int[] arr = { 1, 0, 1, 0, 1 };
int n = arr.Length;
Console.Write(countWays(arr, n));
}
}
// This code is contributed
// by Akanksha Rai
PHP
输出:
4