给定一个整数数组arr [] ,任务是计算可能的分区数量,以使每个分区中的最小元素划分该分区的所有其他元素。分区不必是连续的。
例子:
Input: arr[] = {10, 7, 20, 21, 13}
Output: 3
The possible partitions are {10, 20}, {7, 21} and {13}.
In each partition, all the elements are divisible by
the minimum element of the partition.
Input: arr[] = {7, 6, 5, 4, 3, 2, 2, 3}
Output: 4
方法:
- 在数组中找到不等于INT_MAX的最小元素。
- 从最小元素可整除的数组中删除所有元素(用INT_MAX替换)。
- 由于运算而产生的有效最小元素数是所需的分区数。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the count partitions
// possible from the given array such that
// the minimum element of any partition
// divides all the other elements
// of that partition
int countPartitions(int A[], int N)
{
// Initialize the count variable
int count = 0;
for (int i = 0; i < N; i++) {
// Find the minimum element
int min_elem = *min_element(A, A + N);
// Break if no minimum element present
if (min_elem == INT_MAX)
break;
// Increment the count if
// minimum element present
count++;
// Replace all the element
// divisible by min_elem
for (int i = 0; i < N; i++) {
if (A[i] % min_elem == 0)
A[i] = INT_MAX;
}
}
return count;
}
// Driver code
int main()
{
int arr[] = { 7, 6, 5, 4, 3, 2, 2, 3 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << countPartitions(arr, N);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
static int INT_MAX = Integer.MAX_VALUE ;
static int min_element(int []A, int N)
{
int min = A[0];
int i;
for( i = 1; i < N ; i++)
{
if (min > A[i])
{
min = A[i];
}
}
return min;
}
// Function to return the count partitions
// possible from the given array such that
// the minimum element of any partition
// divides all the other elements
// of that partition
static int countPartitions(int []A, int N)
{
// Initialize the count variable
int count = 0;
int i, j;
for (i = 0; i < N; i++)
{
// Find the minimum element
int min_elem = min_element(A, N);
// Break if no minimum element present
if (min_elem == INT_MAX)
break;
// Increment the count if
// minimum element present
count++;
// Replace all the element
// divisible by min_elem
for (j = 0; j < N; j++)
{
if (A[j] % min_elem == 0)
A[j] = INT_MAX;
}
}
return count;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 7, 6, 5, 4, 3, 2, 2, 3 };
int N = arr.length;
System.out.println(countPartitions(arr, N));
}
}
// This code is contributed by AnkitRai01
Python3
# Python3 implementation of the approach
import sys
INT_MAX = sys.maxsize;
# Function to return the count partitions
# possible from the given array such that
# the minimum element of any partition
# divides all the other elements
# of that partition
def countPartitions(A, N) :
# Initialize the count variable
count = 0;
for i in range(N) :
# Find the minimum element
min_elem = min(A);
# Break if no minimum element present
if (min_elem == INT_MAX) :
break;
# Increment the count if
# minimum element present
count += 1;
# Replace all the element
# divisible by min_elem
for i in range(N) :
if (A[i] % min_elem == 0) :
A[i] = INT_MAX;
return count;
# Driver code
if __name__ == "__main__" :
arr = [ 7, 6, 5, 4, 3, 2, 2, 3 ];
N = len(arr);
print(countPartitions(arr, N));
# This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
static int INT_MAX = int.MaxValue ;
static int min_element(int []A, int N)
{
int min = A[0];
int i;
for( i = 1; i < N ; i++)
{
if (min > A[i])
{
min = A[i];
}
}
return min;
}
// Function to return the count partitions
// possible from the given array such that
// the minimum element of any partition
// divides all the other elements
// of that partition
static int countPartitions(int []A, int N)
{
// Initialize the count variable
int count = 0;
int i, j;
for (i = 0; i < N; i++)
{
// Find the minimum element
int min_elem = min_element(A, N);
// Break if no minimum element present
if (min_elem == INT_MAX)
break;
// Increment the count if
// minimum element present
count++;
// Replace all the element
// divisible by min_elem
for (j = 0; j < N; j++)
{
if (A[j] % min_elem == 0)
A[j] = INT_MAX;
}
}
return count;
}
// Driver code
public static void Main()
{
int []arr = { 7, 6, 5, 4, 3, 2, 2, 3 };
int N = arr.Length;
Console.WriteLine(countPartitions(arr, N));
}
}
// This code is contributed by AnkitRai01
Javascript
输出:
4
时间复杂度: O(N 2 )
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