Overview

Given the initial battery level and per-second consumption rate of each smartphone, find the total remaining battery across all smartphones at time \(T\).

Analysis

Each smartphone \(i\) has an initial battery level \(A_i\) and consumes \(B_i\) units of battery per second. At time \(T\), the battery level would naively be \(A_i - B_i \times T\), but it cannot go negative — the minimum is \(0\).

In other words, the remaining battery of smartphone \(i\) at time \(T\) can be expressed as:

\[ \max(0, A_i - B_i \times T) \]

Therefore, we compute this for all smartphones and sum the results to get the answer.

A straightforward approach of performing this calculation for each smartphone is sufficient, and no particularly complex processing is needed. However, since the constraints are as large as \(N \leq 2 \times 10^5\) and \(T \leq 10^9\), efficient input handling is necessary. Using Python’s standard input() may result in TLE, so a faster input method (e.g., sys.stdin.read) should be used.

Algorithm

1. Read all input at once (for speed).
2. For each smartphone, read the initial battery level \(A_i\) and consumption rate \(B_i\), and perform the following calculation:
   • Consumption at time \(T\): \(B_i \times T\)
   • Remaining battery: \(\max(0, A_i - B_i \times T)\)
3. Sum the remaining battery of all smartphones and output the result.

Complexity

• Time complexity: \(O(N)\)
• Space complexity: \(O(1)\) (excluding input data)

Implementation Notes

• sys.stdin.read is used for fast input.

• max(0, ...) is used to ensure each smartphone’s remaining battery does not go negative.

• Be careful about integer overflow (\(B_i \times T\) can be on the order of \(10^{18}\), but this is not an issue in Python).

  Source Code

import sys
input = sys.stdin.read

def main():
    data = input().split()
    N = int(data[0])
    T = int(data[1])
    
    total = 0
    index = 2
    for _ in range(N):
        A = int(data[index])
        B = int(data[index+1])
        index += 2
        
        # 消費されるバッテリー量
        consumed = B * T
        
        # 残量は0未満にならない
        remaining = max(0, A - consumed)
        total += remaining
    
    print(total)

if __name__ == "__main__":
    main()

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