Overview

This is a problem where, for \(N\) types of part-time jobs, you calculate “hourly wage \(\times\) working hours” for each and find the total sum.

Analysis

The salary earned from each part-time job can be calculated as follows: - 1st part-time job: \(A_1 \times T_1\) yen - 2nd part-time job: \(A_2 \times T_2\) yen - … - \(N\)-th part-time job: \(A_N \times T_N\) yen

The sum of all these values is the total salary Takahashi receives this month. Expressed as a formula, the answer is \(\sum_{i=1}^{N} (A_i \times T_i)\).

Constraints and Notes

• Computational complexity: Since \(N\) is at most \(10^5\), computing and summing each product one by one \(N\) times is well within the time limit.
• Magnitude of values: The maximum possible answer is approximately \(N \times A_i \times T_i = 10^5 \times 10^6 \times 10^6 = 10^{17}\). Although this is a very large number, Python natively supports arbitrary-precision integers (a type that can handle arbitrarily large numbers), so you can compute this directly without any special handling and still get the correct answer.

Algorithm

1. Initialize a variable total_salary to \(0\) to hold the total amount.
2. Read \(N\) from the input.
3. Repeat the following process \(N\) times (\(i = 1, 2, \dots, N\)):
   • Read the hourly wage \(A_i\) and working hours \(T_i\) for the \(i\)-th part-time job.
   • Calculate \(A_i \times T_i\) and add the result to total_salary.
4. Output the final value of total_salary.

Complexity

• Time complexity: \(O(N)\) Since we perform one multiplication and one addition for each of the \(N\) part-time jobs, the processing time is proportional to \(N\).
• Space complexity: \(O(N)\) In this implementation, we use sys.stdin.read().split() to store all input data into a list at once, so the memory usage is proportional to the number of inputs \(N\).

Implementation Notes

• Fast input: Since \(N\) can be as large as \(10^5\), reading all input at once using sys.stdin.read().split() is faster than calling input() \(N\) times.

• Index management: When reading input all at once, the 0th element of the list is \(N\), and from the 1st element onward, the values are arranged as \(A_1, T_1, A_2, T_2, \dots\). It is important to correctly specify the position of the \(i\)-th part-time job’s information in the list using an expression like 2 * i + 1.

  Source Code

import sys

def main():
    # 標準入力からすべてのデータを読み込み、空白文字で分割します
    input_data = sys.stdin.read().split()
    
    # 入力が空の場合は終了します
    if not input_data:
        return
    
    # 最初の値はアルバイトの種類数 N です
    n = int(input_data[0])
    
    total_salary = 0
    # 各アルバイトの時給 A_i と 働いた時間 T_i を順に処理します
    for i in range(n):
        # input_data[1] から順に A1, T1, A2, T2, ... と並んでいます
        a = int(input_data[2 * i + 1])
        t = int(input_data[2 * i + 2])
        # 給料を加算します
        total_salary += a * t
        
    # 合計金額を出力します
    print(total_salary)

if __name__ == '__main__':
    main()

This editorial was generated by gemini-3-flash-thinking.