治愈 N 人所需的最少天数

给定一个数组arr[] ，代表医院中N人的年龄，并且只有一名医生每天最多可以给P人免疫剂量，任务是找到给予免疫所需的最少天数治愈，使得高危人群和正常人不会在同一天服用一剂。
注意：任何年龄≤ 10岁或≥ 60岁的人都被视为高危人群。

例子：

Input: arr[] = {9, 80, 27, 72, 79}, P = 2
Output: 3
Explanation:
There are 4 high risk persons. Therefore, on days 1 and 2, 2 high risk persons can be immuned. On the last day, the normal person can be immuned. Therefore, 3 days are required.

处理方法：按照以下步骤解决问题：

1. 初始化一个变量，比如risk ，以存储年龄小于或等于10且大于或等于60 的人数。
2. 初始化一个变量，比如normal_risk ，以存储年龄在[11, 59]范围内的人数
3. 遍历数组并检查年龄是否小于等于10或大于等于60 。如果发现为真，则增加risk的值。
4. 否则，增加normal_risk的值。
5. 最后，打印ceil(risk / P) + ceil(normal_risk / P) 的值

下面是上述方法的实现：

C++

// C++ Program for the above approach
 
#include 
using namespace std;
 
// Function to find minimum count of days required
// to give a cure such that the high risk person
// and risk person does not get a dose on same day.
void daysToCure(int arr[], int N, int P)
{
 
    // Stores count of persons whose age is
    // less than or equal to 10 and
    // greater than or equal to 60.
    int risk = 0;
 
    // Stores the count of persons
    // whose age is in the range [11, 59]
    int normal_risk = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // If age less than or equal to 10
        // or greater than or equal to 60
        if (arr[i] >= 60 || arr[i] <= 10) {
 
            // Update risk
            risk++;
        }
        else {
 
            // Update normal_risk
            normal_risk++;
        }
    }
 
    // Calculate days to cure risk
    // and normal_risk persons
    int days = (risk / P) + (risk % P > 0)
               + (normal_risk / P)
               + (normal_risk % P > 0);
 
    // Print the days
    cout << days;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 9, 80, 27, 72, 79 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Given P
    int P = 2;
 
    daysToCure(arr, N, P);
 
    return 0;
}

Java

// Java Program for the above approach
class GFG
{
  // Function to find minimum count of days required
  // to give a cure such that the high risk person
  // and risk person does not get a dose on same day.
  static void daysToCure(int arr[], int N, int P)
  {
  
    // Stores count of persons whose age is
    // less than or equal to 10 and
    // greater than or equal to 60.
    int risk = 0;
  
    // Stores the count of persons
    // whose age is in the range [11, 59]
    int normal_risk = 0;
  
    // Traverse the array arr[]
    for (int i = 0; i < N; i++)
    {
  
      // If age less than or equal to 10
      // or greater than or equal to 60
      if (arr[i] >= 60 || arr[i] <= 10)
      {
  
        // Update risk
        risk++;
      }
      else
      {
  
        // Update normal_risk
        normal_risk++;
      }
    }
  
    // Calculate days to cure risk
    // and normal_risk persons
    int days = (risk / P) +  (normal_risk / P);
  
    if(risk % P > 0)
    {
      days++;
    }
  
    if(normal_risk % P > 0)
    {
      days++;
    }
  
    // Print the days
    System.out.print(days);
  } 
 
    public static void main(String[] args) {
        // Given array
        int arr[] = { 9, 80, 27, 72, 79 };
      
        // Size of the array
        int N = arr.length;
      
        // Given P
        int P = 2;  
        daysToCure(arr, N, P);
    }
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 Program for the above approach
 
# Function to find minimum count of days required
# to give a cure such that the high risk person
# and risk person does not get a dose on same day.
def daysToCure(arr, N, P):
 
    # Stores count of persons whose age is
    # less than or equal to 10 and
    # greater than or equal to 60.
    risk = 0
 
    # Stores the count of persons
    # whose age is in the range [11, 59]
    normal_risk = 0
 
    # Traverse the array arr[]
    for i in range(N):
 
        # If age less than or equal to 10
        # or greater than or equal to 60
        if (arr[i] >= 60 or arr[i] <= 10):
 
            # Update risk
            risk += 1
        else:
 
            # Update normal_risk
            normal_risk += 1
 
    # Calculate days to cure risk
    # and normal_risk persons
    days = (risk // P) + (risk % P > 0) + (normal_risk // P) + (normal_risk % P > 0)
 
    # Prthe days
    print (days)
 
# Driver Code
if __name__ == '__main__':
   
    # Given array
    arr = [9, 80, 27, 72, 79 ]
 
    # Size of the array
    N = len(arr)
 
    # Given P
    P = 2
 
    daysToCure(arr, N, P)
 
    # This code is contributed by mohit kumar 29.

C#

// C# Program for the above approach
using System;
class GFG
{
 
  // Function to find minimum count of days required
  // to give a cure such that the high risk person
  // and risk person does not get a dose on same day.
  static void daysToCure(int[] arr, int N, int P)
  {
 
    // Stores count of persons whose age is
    // less than or equal to 10 and
    // greater than or equal to 60.
    int risk = 0;
 
    // Stores the count of persons
    // whose age is in the range [11, 59]
    int normal_risk = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++)
    {
 
      // If age less than or equal to 10
      // or greater than or equal to 60
      if (arr[i] >= 60 || arr[i] <= 10)
      {
 
        // Update risk
        risk++;
      }
      else
      {
 
        // Update normal_risk
        normal_risk++;
      }
    }
 
    // Calculate days to cure risk
    // and normal_risk persons
    int days = (risk / P) +  (normal_risk / P);
 
    if(risk % P > 0)
    {
      days++;
    }
 
    if(normal_risk % P > 0)
    {
      days++;
    }
 
    // Print the days
    Console.Write(days);
  }
 
  // Driver code
  static void Main()
  {
 
    // Given array
    int[] arr = { 9, 80, 27, 72, 79 };
 
    // Size of the array
    int N = arr.Length;
 
    // Given P
    int P = 2;  
    daysToCure(arr, N, P);
  }
}
 
// This code is contributed by divyesh072019.

Javascript

// javascript Program for the above approach
 
    // Function to find minimum count of days required
    // to give a cure such that the high risk person
    // and risk person does not get a dose on same day.
    function daysToCure(arr , N , P) {
 
        // Stores count of persons whose age is
        // less than or equal to 10 and
        // greater than or equal to 60.
        var risk = 0;
 
        // Stores the count of persons
        // whose age is in the range [11, 59]
        var normal_risk = 0;
 
        // Traverse the array arr
        for (i = 0; i < N; i++) {
 
            // If age less than or equal to 10
            // or greater than or equal to 60
            if (arr[i] >= 60 || arr[i] <= 10) {
 
                // Update risk
                risk++;
            } else {
 
                // Update normal_risk
                normal_risk++;
            }
        }
 
        // Calculate days to cure risk
        // and normal_risk persons
        var days = parseInt((risk / P) + (normal_risk / P));
 
        if (risk % P > 0) {
            days++;
        }
 
        if (normal_risk % P > 0) {
            days++;
        }
 
        // Print the days
        document.write(days);
    }
 
     
        // Given array
        var arr = [ 9, 80, 27, 72, 79 ];
 
        // Size of the array
        var N = arr.length;
 
        // Given P
        var P = 2;
        daysToCure(arr, N, P);
 
// This code contributed by Rajput-Ji
 

输出

3

时间复杂度： O(N)
辅助空间： O(1)