为矩形公园浇水所需的最少洒水器

给定N * M 个矩形公园，有N行和M列，公园的每个单元格都是一个单位面积的正方形，单元格之间的边界称为六边形，可以在六边形中间放置一个洒水器。任务是找到为整个公园浇水所需的最少洒水器数量。

例子：

Input: N = 3 M = 3
Output: 5
Explanation:
For the first two columns 3 sprinklers are required and for last column we are bound to use 2 sprinklers to water the last column.

Input: N = 5 M = 3
Output: 8
Explanation:
For the first two columns 5 sprinklers are required and for last column we are bound to use 3 sprinklers to water the last column.

方法：

1. 在进行一些观察之后，可以指出一件事，即每两列需要N 个喷水器，因为我们可以将它们放置在两列之间。
2. 如果M是偶数，那么显然需要N* (M / 2) 个喷头。
3. 但是，如果M是奇数，则可以使用偶数列公式计算M – 1列的解决方案，并且对于最后一列添加( N + 1) / 2喷头浇灌最后一列，无论 N 是奇数还是偶数。

   这是上述方法的实现：

   C++

   // C++ program to find the
   // minimum number sprinklers
   // reqired to water the park.
     
   #include 
   using namespace std;
   typedef long long int ll;
     
   // Function to find the
   // minimum number sprinklers
   // required to water the park.
   void solve(int N, int M)
   {
     
       // General requirements of
       // sprinklers
       ll ans = (N) * (M / 2);
     
       // if M is odd then add
       // one additional sprinklers
       if (M % 2 == 1) {
           ans += (N + 1) / 2;
       }
     
       cout << ans << endl;
   }
     
   // Driver code
   int main()
   {
       int N, M;
       N = 5;
       M = 3;
       solve(N, M);
   }

   Java

   // Java program to find minimum 
   // number sprinklers required 
   // to cover the park
   class GFG{ 
         
   // Function to find the minimum 
   // number sprinklers reqired 
   // to water the park. 
   public static int solve(int n, int m)
   {
         
       // General requirements of sprinklers
       int ans = n * (m / 2);
             
       // If M is odd then add one
       // additional sprinklers 
       if (m % 2 == 1)
       {
           ans += (n + 1) / 2;
       } 
       return ans;
   }
     
   // Driver code
   public static void main(String args[]) 
   { 
       int N = 5;
       int M = 3;
         
       System.out.println(solve(N, M)); 
   } 
   } 
     
   // This code is contributed by grand_master

   Python3

   # Python3 program to find the 
   # minimum number sprinklers 
   # required to water the park. 
     
   # Function to find the 
   # minimum number sprinklers 
   # reqired to water the park. 
   def solve(N, M) :
         
       # General requirements of 
       # sprinklers 
       ans = int((N) * int(M / 2))
     
       # if M is odd then add 
       # one additional sprinklers 
       if (M % 2 == 1):
           ans += int((N + 1) / 2)
     
       print(ans)
     
   # Driver code 
   N = 5
   M = 3
   solve(N, M)
     
   # This code is contributed by yatinagg

   C#

   // C# program to find minimum 
   // number sprinklers required 
   // to cover the park
   using System;
     
   class GFG{ 
         
   // Function to find the minimum 
   // number sprinklers reqired 
   // to water the park. 
   public static int solve(int n, int m)
   {
         
       // General requirements of sprinklers
       int ans = n * (m / 2);
             
       // If M is odd then add one
       // additional sprinklers 
       if (m % 2 == 1)
       {
           ans += (n + 1) / 2;
       } 
       return ans;
   }
     
   // Driver code
   public static void Main(String []args) 
   { 
       int N = 5;
       int M = 3;
         
       Console.WriteLine(solve(N, M)); 
   } 
   } 
     
   // This code is contributed by 29AjayKumar

   输出：

   8
   

   时间复杂度： O(1)
   辅助空间： O(1)

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