Overview

Given the compensation amounts \(P_i\) of \(N\) employees, the problem asks to find the total sum of only those amounts that are multiples of a specified value \(K\).

Analysis

The key points for solving this problem are the following two:

1. How to determine multiples Whether a number \(P_i\) is a multiple of \(K\) can be determined in programming by checking “whether the remainder of dividing \(P_i\) by \(K\) is \(0\).” In most languages, this can be written as the conditional expression P_i % K == 0 using the % operator.

2. Computational complexity and magnitude of values

   • The number of employees \(N\) is at most \(2 \times 10^5\). An \(O(N)\) algorithm that checks each employee exactly once is well within the time limit.
   • Since the compensation \(P_i\) can be up to \(10^9\) and the number of employees \(N\) can be up to \(2 \times 10^5\), the total sum can reach approximately \(2 \times 10^{14}\). In Python, large integers (arbitrary-precision integers) are handled automatically, so this is not an issue. However, in languages like C++, a 64-bit integer type (such as long long) must be used.

Algorithm

1. Initialize a variable total_sum to \(0\) to hold the total sum.
2. Check each employee’s compensation \(P_i\) one by one in order.
3. If \(P_i\) is divisible by \(K\) (P_i % K == 0), add that value to total_sum.
4. After checking all compensation amounts, output total_sum.

Complexity

• Time complexity: \(O(N)\) Since we scan through \(N\) elements once each, the computation completes in time proportional to the input size.
• Space complexity: \(O(N)\) In this implementation, all input values are stored in a list (array), resulting in \(O(N)\). Note that if values are processed one by one as they are read, this can be reduced to \(O(1)\).

Implementation Notes

• Fast I/O: Since \(N\) can be large, repeatedly calling input() in Python may result in long execution times. By using sys.stdin.read().split() to read all input at once, we speed up the processing.

• Conditional branching: We use if p % k == 0: to ensure that values are added only when they are multiples.

  Source Code

import sys

def solve():
    input_data = sys.stdin.read().split()
    if not input_data:
        return
    
    n = int(input_data[0])
    k = int(input_data[1])
    p_list = map(int, input_data[2:])
    
    total_sum = 0
    for p in p_list:
        if p % k == 0:
            total_sum += p
            
    print(total_sum)

if __name__ == "__main__":
    solve()

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