Overview

For each of \(N\) types of sweets, this problem asks you to find the number per person (quotient) and the remainder when equally dividing among \(M\) people.

Analysis

This problem can be solved by simply computing the quotient and remainder when dividing each sweet count \(A_i\) by \(M\). From the constraints, each \(A_i\) can be up to \(10^9\) and \(M\) can also be up to \(10^9\), but since integer division and modulo operations execute quickly, even a straightforward approach is sufficiently efficient.

A point worth noting is that even when \(A_i\) is smaller than \(M\), the correct result of quotient 0 and remainder \(A_i\) is obtained. Similarly, when \(A_i\) is 0, the correct result of quotient 0 and remainder 0 is produced.

Algorithm

For each \(A_i\), perform the following calculations: - Quotient \(Q_i = A_i // M\) (integer division) - Remainder \(R_i = A_i \% M\) (modulo operation)

This computation satisfies the mathematical definition \(A_i = M \times Q_i + R_i\) (where \(0 \leq R_i < M\)). Since Python’s integer division and modulo operations automatically satisfy this condition, no special handling is needed.

Complexity

• Time complexity: \(O(N)\)
  • Because the quotient and remainder are computed in constant time for each of the \(N\) sweets
• Space complexity: \(O(N)\)
  • Because the counts of the \(N\) sweets from the input need to be stored

Implementation Notes

• Use sys.stdin.read().split() for efficient input reading

• For each \(A_i\), compute the quotient with the // operator and the remainder with the % operator

• Output is expressed concisely using the format string f"{q} {r}"

• Since Python’s integer arithmetic handles large numbers accurately, there is no need to worry about overflow

  Source Code

import sys

def main():
    data = sys.stdin.read().split()
    n = int(data[0])
    m = int(data[1])
    A = list(map(int, data[2:2+n]))
    
    for a in A:
        q = a // m
        r = a % m
        print(f"{q} {r}")

if __name__ == "__main__":
    main()

This editorial was generated by deepseekv3.