Overview

This is a problem where you calculate “hourly wage × hours” for each part-time job and sum them all up to find the total salary for the month. In other words, you simply need to compute \( \sum_{i=1}^{N} A_i T_i \).

Analysis

The essence of this problem is very simple. The pay for each job is given by: \(A_i\) (hourly wage) \(\times\) \(T_i\) (hours worked)

Therefore, the total salary is the following sum:

\[ A_1T_1 + A_2T_2 + \cdots + A_NT_N \]

Key Observations

• The information for each part-time job is independent, so you can read one line at a time and accumulate the sum as you go.
• There is no need to store everything in an array; you can add to the total on the spot.

Notes on the Straightforward Approach

No special tricks are needed for this problem. Since \(N \le 10^5\), computing each term one at a time is fast enough. In fact, unnecessarily complex processing (such as nested loops) could make the solution slower for no reason.

Also, considering the maximum values: \(A_iT_i \le 10^6 \times 10^6 = 10^{12}\), and summing up to \(10^5\) such values means the total can be at most around \(10^{17}\). Python’s int can handle large integers, so there is no issue.

Algorithm

1. Read \(N\).
2. Initialize the total total to 0.
3. Repeat \(N\) times:
   • Read \(A, T\)
   • Perform total += A * T
4. Output total.

Complexity

• Time complexity: \(O(N)\)
• Space complexity: \(O(1)\)

Implementation Notes

• Since there are many input lines, using input = sys.stdin.readline is faster and more stable.

• By accumulating the sum incrementally without storing values in an array, memory usage is kept low.

• In Python, you can compute the products and total sum without worrying about overflow.

  Source Code

import sys

def main():
    input = sys.stdin.readline
    N = int(input())
    total = 0
    for _ in range(N):
        A, T = map(int, input().split())
        total += A * T
    print(total)

if __name__ == "__main__":
    main()

This editorial was generated by gpt-5.3-codex.