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A. Race time limit per test2 seconds memory limit ...

Creato il: 23 giugno 2025

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Domanda

A. Race
time limit per test2 seconds
memory limit per test256 megabytes
Alice and Bob participate in a game TV show. When the game starts, the prize will be dropped to a certain point, and whoever gets to it first will get the prize.

Alice decided that she would start running from point a
. Bob, however, has not yet chosen his starting position.

Bob knows that the prize could drop either at point x
or at point y
. He also knows that he can reach the prize faster than Alice if the distance from his starting position to the prize is strictly less than the distance from Alice's starting position to the prize. The distance between any two points c
and d
is calculated as |c−d|
.

Your task is to determine whether Bob can choose an integer point that is guarantee to get to the prize faster, regardless of where it appears (at point x
or y
). Bob can choose any integer point, except for a
(in particular, he can choose to start in point x
, point y
, or any other point, but not a
).

Input
The first line contains a single integer t
(1≤t≤1000
) — the number of test cases.

The only line of each test case contains three integers a
, x
, y
(1≤a,x,y≤100
). Points a
, x
, and y
are pairwise distinct.

Output
For each test case, print "YES" (case insensitive) if Bob can choose an integer point that is guarantee to get to the prize faster, regardless of where it appears. Otherwise, print "NO" (case insensitive).

Example
InputCopy
3
1 3 4
5 3 1
3 1 5
OutputCopy
YES
YES
NO
Note
In the first example, Bob can choose point 4
. If the prize will be at point x
, then Bob's distance is |4−3|=1
and Alice's distance is |1−3|=2
. If the prize will be at point y
, then Bob's distance is |4−4|=0
and Alice's distance is |1−4|=3
.

In the second example, Bob can choose point 2
. If the prize will be at point x
, then Bob's distance is |2−3|=1
and Alice's distance is |5−3|=2
. If the prize will be at point y
, then Bob's distance is |2−1|=1
and Alice's distance is |5−1|=4
.

In the third example, Bob cannot choose a point to guarantee his victory.

c++ code

fix this code

#include<bits/stdc++.h>
using namespace std;

int main() {
int t;
cin >> t;
while (t--) {
int p, m, n;
cin >> p >> m >> n;

text
    int d1 = abs(p - m);
    int d2 = abs(p - n);
    int d = abs(m - n);
    int mn = d1 < d2 ? d1 : d2;

    cout << (d < 2 * mn ? "YES" : "NO") << endl;

}
return 0;

}

Pensiero

Analyzing race problem

The problem involves Alice at point a and Bob needing to choose a starting point b (where b ≠ a). Bob wins if his distance to both potential prize drop locations (x, y) is strictly less than Alice's. We need to check for a valid b such that the constraints abs(b-x) < abs(a-x) and abs(b-y) < abs(a-y) hold. Then we output YES if such a b exists; otherwise, NO. The user seems to need help with fixing their C++ code to implement this logic effectively.

Reviewing code logic

The user provided code that attempts to check distances between points, but it seems the logic may not be correct. The goal is to find a point b (not equal to a) where the distance from b to both x and y is strictly less than Alice’s distance from x and y. The correct approach involves defining valid intervals based on Alice's position and checking if there’s an overlap that doesn't include Alice's position. I think overlaps occur if the ranges are valid and don't exclusively cover Alice's position. If they do, no valid b exists.

Fixing C++ code logic

The C++ code provided by the user uses incorrect conditions for checking whether Bob can guarantee a win. The issue stems from their use of d < 2*mn, which doesn't correctly check the intersection of intervals. The correct approach involves checking if the distance between prize locations is less than the sum of Alice's and Bob's distances minus 2. This ensures that valid intervals exist for Bob to win. I'll need to revise the code to handle these correct conditions for determining the answer.

Explaining the bug fix

The user needs a fixed C++ code to accurately check if a specific point a can be chosen for a given range defined by x and y. The scenario involves checking if two intervals overlap, with one exception — the chosen value cannot be a. I’ve worked out that the correct condition for determining overlap involves verifying whether the difference between x and y is less than or equal to the sum of distances, subtracting 2. I’ve highlighted this critical point and detailed the necessary adjustments in the code. Now it's ready to properly reflect those changes.

Thought for 44 seconds

Risposta

The mistake is that Bob’s “winning” starting positions form the intersection of two intervals:

text
I₁ = { b : |b−x| < |a−x| }  ⇔  x−D₁+1 ≤ b ≤ x+D₁−1
I₂ = { b : |b−y| < |a−y| }  ⇔  y−D₂+1 ≤ b ≤ y+D₂−1

where D₁=|a−x| and D₂=|a−y|. That intersection is non‐empty exactly when

|x−y| ≤ D₁ + D₂ − 2.

Below is a corrected C++ implementation:

cpp
#include <bits/stdc++.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;
    while (t--) {
        int a, x, y;
        cin >> a >> x >> y;

        int D1 = abs(a - x);
        int D2 = abs(a - y);
        int distXY = abs(x - y);

        // Intersection non-empty ⇔ distXY <= D1 + D2 - 2
        if (distXY <= D1 + D2 - 2) {
            cout << "YES\n";
        } else {
            cout << "NO\n";
        }
    }

    return 0;
}

Explanation:

  • Let Alice start at a and the prize be at either x or y.
  • Bob wants a point b such that
      • |b−x| < |a−x| and
      • |b−y| < |a−y|.
  • Each of these is an interval of length 2·D₁−1 and 2·D₂−1 around x and y respectively.
  • Two such intervals overlap if and only if 
    text
    |x−y| ≤ (D₁−1) + (D₂−1)
       = D₁ + D₂ − 2.
  • If they do overlap, that overlap is comprised of integer points b ≠ a, so Bob can pick one and always beat Alice.

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