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code in cpp

作成日: 2025年7月12日

使用モデル O3 by Chat01

質問

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47e5930e-68a8-4eb2-bf2d-46aa496ca2b7.jpg47e5930e-68a8-4eb2-bf2d-46aa496ca2b7.jpg

code in cpp

Python

python
from PIL import Image
import matplotlib.pyplot as plt

img1 = Image.open('/mnt/data/a5431805-e290-43b0-87f4-4d0397195695.jpg')
img1.size

(720, 1280)

Python

python
plt.imshow(img1)
plt.axis('off')

(-0.5, 719.5, 1279.5, -0.5)

Python

python
img2 = Image.open('/mnt/data/e3f20efe-ea8e-4237-8749-2cb85ed0d1e7.jpg')
plt.imshow(img2); plt.axis('off')

(-0.5, 719.5, 1279.5, -0.5)

Python

python
plt.figure(figsize=(6,10))
plt.imshow(img2.crop((0,0,720,640)))
plt.axis('off')

(-0.5, 719.5, 639.5, -0.5)

Python

python
img4 = Image.open('/mnt/data/47e5930e-68a8-4eb2-bf2d-46aa496ca2b7.jpg')
plt.imshow(img4); plt.axis('off')

(-0.5, 719.5, 1279.5, -0.5)

Python

python
plt.figure(figsize=(6,10))
plt.imshow(img4.crop((0,0,720,700)))
plt.axis('off')

(-0.5, 719.5, 699.5, -0.5)

Python

python
plt.figure(figsize=(6,10))
plt.imshow(img2.crop((0,640,720,1280)))
plt.axis('off')

(-0.5, 719.5, 639.5, -0.5)

Python

python
img3 = Image.open('/mnt/data/9c8c0d8f-a19c-4a6c-8ea4-4de86f257692.jpg')
plt.imshow(img3); plt.axis('off')

(-0.5, 719.5, 1279.5, -0.5)

Python

python
plt.figure(figsize=(7,10))
plt.imshow(img3.crop((0,0,720,500)))
plt.axis('off')

(-0.5, 719.5, 499.5, -0.5)

Python

python
plt.figure(figsize=(7,10))
plt.imshow(img4.crop((0,0,720,400)))
plt.axis('off')

(-0.5, 719.5, 399.5, -0.5)

Python

python
plt.figure(figsize=(7,10))
plt.imshow(img4.crop((0,350,720,750)))
plt.axis('off')

(-0.5, 719.5, 399.5, -0.5)

Python

python
plt.figure(figsize=(4,8))
plt.imshow(img3.crop((100,400,400,780)))
plt.axis('off')

(-0.5, 299.5, 379.5, -0.5)

Python

python
plt.imshow(img2.crop((100,400,400,780))); plt.axis('off')

(-0.5, 299.5, 379.5, -0.5)

思考

Clarifying Bank and Stub Setup

The input sample mentions 5 logs (1, 6, 5, 4, 1), and although the left and right banks include a stub of 1, it seems that only the logs are counted. This suggests the starting position is not the bank's stub but actually the vertical start at the bank. From there, the first jump goes straight down to log I. The horizontal move from the bank to log I may be negligible, but the difference between horizontal positions could be 1 based on the setup.

Refining Jump Algorithm

The bank's capacity isn't given, so I'll assume the first jump is always possible. Similarly, the final jump to the opposite bank should be checked based on the capacity. The number "1" on the opposite bank might signify the final step, influencing whether the last jump is within capacity. The algorithm checks each pair of logs for the difference, and if the jump is too large, it fails. When failure occurs, the number of jumps is the forward and then back. Sample diff calculations confirm the approach by matching the expected return on total jumps.

Thought for 2m 40s

回答

Below is a compact O(N) solution that follows exactly what the statement (and the diagrams) describe.

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

/*
 * n  – number of logs
 * x  – max steps she can walk **left** on a log
 * y  – max steps she can walk **right** on a log
 * a  – lengths of the logs, in the order she encounters them
 *
 * Returns the total number of jumps Geeta makes before she either
 *   • reaches the opposite bank, or
 *   • is forced to turn around and return to the start.
 */
long long stepsNeeded(int n, long long x, long long y,
                      const vector<long long>& a)
{
    long long jumps = 1;          // first jump: start bank → log 0

    for (int i = 0; i < n - 1; ++i)
    {
        long long diff = a[i + 1] - a[i];   // horizontal walk on log i

        if (diff > 0)           // need to walk to the **right**
        {
            if (diff > y)              // beyond her capability → must retreat
                return jumps * 2;      // go back the way she came
        }
        else if (diff < 0)      // need to walk to the **left**
        {
            if (-diff > x)              // beyond her capability → must retreat
                return jumps * 2;
        }
        ++jumps;                // successful jump to log (i+1)
    }

    // All logs cleared – one final jump to the far bank
    return jumps + 1;
}

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

    int n; long long x, y;
    if (!(cin >> n >> x >> y)) return 0;

    vector<long long> a(n);
    for (int i = 0; i < n; ++i) cin >> a[i];

    cout << stepsNeeded(n, x, y, a) << '\n';
    return 0;
}

How the algorithm matches the story

  1. Forward traversal
    For every pair of consecutive logs we compute how far Geeta has to shuffle horizontally on the current log to align with the next log’s right-end:
    diff = a[i+1] - a[i]

    • diff > 0 → walk right diff steps (must be ≤ y).
    • diff < 0 → walk left |diff| steps (must be ≤ x).

    If the walk is within her capability she performs the jump (so jumps++) and continues; otherwise she cannot progress.

  2. Retreat rule
    The moment a required horizontal walk exceeds her limit, she turns back.
    All jumps she made so far are perfectly reversible (they already satisfied the limits in the opposite direction), so the retreat costs exactly the same number of jumps.
    Hence we return 2 × (current forward jumps).

  3. Success
    If every gap satisfies the limits, she stands on the last log; one extra leap gets her onto the far bank, so we add +1.

Sample 1
N=5, X=3, Y=5, a=[1,6,5,4,1]
All horizontal walks (5 R, 1 L, 1 L, 3 L) fit the limits ⇒ total jumps = 5 logs + 1 = 6.

Sample 2
N=5, X=1, Y=5, a=[1,6,5,4,1]
Fourth gap needs 3 L > X, so she retreats:
forward jumps = 4 → answer = 2 × 4 = 8.

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