Overview

This problem asks us to find the total intensity of sound reaching a measurement point \(P\) from \(N\) speakers placed on a number line.

Analysis

As stated in the problem description, the sound intensity reaching the measurement point from the \(i\)-th speaker is represented as \(\frac{V_i}{|X_i - P|}\). If a speaker is located at the exact same coordinate as the measurement point (\(X_i = P\)), division by zero occurs, so it must be excluded from the calculation.

No special techniques or advanced algorithms are required; a straightforward approach of checking the condition and summing up the intensities for each speaker sequentially is sufficient. Given the constraints \(N \leq 2 \times 10^5\), an \(O(N)\) calculation processing one speaker at a time will comfortably run within the time limit.

Algorithm

1. Input the number of speakers \(N\) and the coordinate of the measurement point \(P\).
2. Initialize a variable (floating-point type) to hold the total sum with \(0\).
3. For each speaker, input its coordinate \(X_i\) and output intensity \(V_i\), then perform the following:
   • If \(X_i \neq P\), calculate the distance \(d = |X_i - P|\).
   • Add \(\frac{V_i}{d}\) to the total sum variable.
   • If \(X_i = P\), do nothing (exclude it).
4. Finally, output the total sum. Since the acceptable error is \(10^{-4}\), output with a sufficient number of decimal places.

Complexity

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

Implementation Details

• The coordinates \(P\) and \(X_i\) are integers between \(-10^9\) and \(10^9\), and the difference \(|X_i - P|\) can be up to \(2 \times 10^9\). Therefore, you should use a 64-bit integer type (long long) instead of a 32-bit integer type (int).

• When performing division, cast either the dividend or the divisor to a floating-point number (double) to avoid integer division.

• When printing the result, to satisfy the required precision (absolute error of at most \(10^{-4}\)), it is safer to specify a larger number of decimal places in your output.

  Source Code

#include <iostream>
#include <iomanip>

using namespace std;

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int N;
    long long P;
    if (!(cin >> N >> P)) return 0;
    
    double ans = 0.0;
    for (int i = 0; i < N; i++) {
        long long X, V;
        cin >> X >> V;
        if (X != P) {
            long long dist = (X > P) ? (X - P) : (P - X);
            ans += static_cast<double>(V) / dist;
        }
    }
    
    cout << fixed << setprecision(10) << ans << "\n";
    return 0;
}

This editorial was generated by or-glm-5.2-high.