Sum of count of persons still performing the job whenever a person finishes job

Last Updated : 18 Nov, 2021

Given two integers P denoting the number of people and a positive integer K. Each person is assigned with the same job which takes exactly K hours to finish. A condition is given that all of them can start performing their jobs exactly in X hours of interval. Count the number of persons still performing the job whenever a prior person finishes his/her job. The task is to find the sum of such counts.

Examples:

Input: P = 4, K = 6, X = 3
Output: 5
Explanation: Let the four persons be P₁, P₂, P₃, P₄

P₁ starts at 0 and finishes at 6

P₂ starts at 3 and finishes at 9

P₃starts at 6 and finishes at 12

P₄ starts at 9 and finishes at 15

So, when P₁ finishes, P₂and P₃ started performing their respective job, count for P₁ = 2
when P₂ finishes, P₃and P₄started performing their respective job, count for P₂ = 2
when P₃ finishes, only P₄ started performing the job, count for P₃ = 1
when P₄ finishes, there is no person who starts at this point, so count for P₄= 0
Therefore, Total counts = (2 + 2 + 1 + 0) = 5

Input: P = 9, K = 72, X = 8
Output: 36

Approach: The given problem can be solved by analyzing the problem with the concept of math. Follow the steps below to solve the problem:

Find the minimum of total persons excluding the first one(because P₁ always starts at 0) and K/X, store it in a variable say a.
Check if a is equal to 0
- if yes, return 0.
- else, calculate the total sum of count by making mathematical formula.
Return the final sum of count as the required answer.

Below is the implementation of the above approach:

C++

// C++ implementation for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the sum of count
// of persons that are still
// performing the job
int countPersons(int P, int K, int X)
{
    // Find the minimum of total persons
    // excluding the first one and K/X
    int a = min(P - 1, K / X);
 
    // If a is equal to 0, return 0
    if (a == 0) {
        return 0;
    }
 
    // Else calculate the total sum of
    // count by the making a formula
    int ans = max(0,
                  a * (a - 1) / 2)
              + a * (P - a);
 
    // Return the sum of count
    return ans;
}
 
// Driver Code
int main()
{
    int P = 22, K = 9, X = 1;
 
    cout << countPersons(P, K, X);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG {
 
// Function to find the sum of count
// of persons that are still
// performing the job
static int countPersons(int P, int K, int X)
{
    // Find the minimum of total persons
    // excluding the first one and K/X
    int a = Math.min(P - 1, K / X);
 
    // If a is equal to 0, return 0
    if (a == 0) {
        return 0;
    }
 
    // Else calculate the total sum of
    // count by the making a formula
    int ans = Math.max(0,
                  a * (a - 1) / 2)
              + a * (P - a);
 
    // Return the sum of count
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    int P = 22, K = 9, X = 1;
    System.out.println(countPersons(P, K, X));
}
}
// This code is contributed by Samim Hossain Mondal.

Python3

# Python program for the above approach
 
# Function to find the sum of count
# of persons that are still
# performing the job
def countPersons(P, K, X):
 
    # Find the minimum of total persons
    # excluding the first one and K/X
    a = min(P - 1, K / X)
 
    # If a is equal to 0, return 0
    if a == 0:
        return 0
 
    # Else calculate the total sum of
    # count by the making a formula
    ans = max(0,
              a * (a - 1) / 2)
 
    # Return the sum of count
    return ans + a * (P - a)
 
# Driver Code
if __name__ == "__main__":
    P = 22
    K = 9
    X = 1
    print(int(countPersons(P, K, X)))
 
# This code is contributed by Potta Lokesh

C#

// C# program for the above approach
using System;
class GFG {
 
// Function to find the sum of count
// of persons that are still
// performing the job
static int countPersons(int P, int K, int X)
{
    // Find the minimum of total persons
    // excluding the first one and K/X
    int a = Math.Min(P - 1, K / X);
 
    // If a is equal to 0, return 0
    if (a == 0) {
        return 0;
    }
 
    // Else calculate the total sum of
    // count by the making a formula
    int ans = Math.Max(0,
                  a * (a - 1) / 2)
              + a * (P - a);
 
    // Return the sum of count
    return ans;
}
 
// Driver Code
public static void Main()
{
    int P = 22, K = 9, X = 1;
    Console.Write(countPersons(P, K, X));
}
}
// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
// Javascript implementation for the above approach
 
// Function to find the sum of count
// of persons that are still
// performing the job
function countPersons(P, K, X)
{
    // Find the minimum of total persons
    // excluding the first one and K/X
    let a = Math.min(P - 1, K / X);
 
    // If a is equal to 0, return 0
    if (a == 0) {
        return 0;
    }
 
    // Else calculate the total sum of
    // count by the making a formula
    let ans = Math.max(0,
                  a * (a - 1) / 2)
              + a * (P - a);
 
    // Return the sum of count
    return ans;
}
 
// Driver Code
let P = 22, K = 9, X = 1;
document.write(countPersons(P, K, X));
 
// This code is contributed by Samim Hossain Mondal.
</script>