Overview

This problem asks you to compute each participant’s “skill value” \(H_i + S_i\) and find the participant number whose value is the maximum (\(D_i\) is ignored).

Analysis

• The total score is \(H_i + S_i + D_i\), but what Aoki wants to see is the pure skill excluding the bonus, so the value to compare is the skill value \(H_i + S_i\).
  \(D_i\) is given as input, but it is irrelevant to determining the winner in this problem.
• All we need to do is “find the person with the maximum \(H_i + S_i\)”, so it suffices to look at each person once and update the maximum.
• A straightforward approach of “storing all skill values in an array and finding the maximum at the end” also works, but there’s no need to store them in an array — we can update the maximum while reading input, saving memory as well (since \(N \le 2\times 10^5\), either approach is fast enough, but the latter is more concise and safe).
• It is guaranteed that exactly one participant has the maximum, so there is no need to handle ties (multiple people sharing the maximum).

Example:
With 3 participants:
1: \((H,S,D)=(10,20,100)\) → skill value \(30\)
2: \((5,40,0)\) → skill value \(45\)
3: \((30,10,1000)\) → skill value \(40\)
The maximum is \(45\) for participant 2, so the answer is 2 (even though \(D\) is large, it doesn’t matter).

Algorithm

1. Initialize best_score (the maximum skill value so far) to \(-1\).
2. Prepare best_idx (the participant number at that point).
3. For \(i=1\) to \(N\), read the input in order and compute the skill value score = h + s.
4. If score > best_score, update best_score and best_idx.
5. Output best_idx at the end.

Complexity

• Time complexity: \(O(N)\) (just looking at each person once)
• Space complexity: \(O(1)\) (only storing the maximum value and the index)

Implementation Notes

• Participant numbers are 1-indexed as stated in the problem, so the loop should be for i in range(1, N + 1):.

• Since there can be up to \(2\times 10^5\) lines of input, using sys.stdin.readline in Python ensures stable and fast reading.

• \(H_i, S_i\) are at most \(10^9\), but their sum is at most \(2\times 10^9\), which Python’s int can handle safely.

  Source Code

import sys

def main():
    input = sys.stdin.readline
    N = int(input())
    best_score = -1
    best_idx = -1
    for i in range(1, N + 1):
        h, s, d = map(int, input().split())
        score = h + s
        if score > best_score:
            best_score = score
            best_idx = i
    print(best_idx)

if __name__ == "__main__":
    main()

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This editorial was generated by gpt-5.2-high.