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solve in cpp

Criado em: 5 de julho de 2025

Respondido usando O3 por Chat01

Pergunta

image.pngimage.png
image.pngimage.png

solve in cpp

Python

python
from PIL import Image
import pytesseract

img = Image.open('/mnt/data/image.png')
img

<PIL.PngImagePlugin.PngImageFile image mode=RGB size=946x914>

Python

python
import pytesseract
text = pytesseract.image_to_string(img)
print(text[:2000])

Python

python
# crop middle portion where sample input text appears; coordinates approximate from screenshot size 946x914
w,h=img.size
crop = img.crop((0,300, w, 600))
crop

<PIL.Image.Image image mode=RGB size=946x300>

Python

python
print(pytesseract.image_to_string(crop))

Python

python
crop2 = img.crop((0,0, w, 250))
crop2

<PIL.Image.Image image mode=RGB size=946x250>

Python

python
print(text)

Python

python
# bottom area
crop3 = img.crop((0,600, w, h))
crop3

<PIL.Image.Image image mode=RGB size=946x314>

Pensamento

Clarifying starting index rule

It seems like the screenshot only shows sample 1, and sample 0 might have been cropped or not captured. Based on the explanation, we start at the 4th bank, but the exact rule for the starting index is still unclear. I thought it might be related to a formula like (n - k) % n, but in Example 0 with n=3 and k=2, the starting index was 0, not 1. So maybe the right formula is (k % n)? I’ll reconcile this further.

Identifying the starting index

I wondered if the algorithm starts at the bank with the most candies, but the examples didn't support that. I tested a few scenarios, and it seems the correct rule might be to start at a bank indexed with (k mod n) — but that didn’t work consistently.

Perhaps the algorithm depends on the parity of k, stepping forward for even k and backward for odd k. It also seems like the goal might be to maximize the total candies by starting at the most optimal index.

Maximizing candies with steps

I realized the starting index rr determines which indices are visited with steps of size kk. For example, in one problem, after computing the suffix sums, I found that starting at index 4 yielded the most total candies, which made sense.

So, the real task is to calculate the possible sums from each starting index (from 0 to n−1n-1) and find the one that maximizes the total candies, adding in the extra candies cc at the end. The output is just the result.

Thought for 1m 36s

Resposta

Idea

Treat each bank i as a possible starting point.
From that start Maku keeps jumping exactly k positions until she runs past the last bank (i, i+k, i+2k, … < n).
The sum of candies collected on that path is

pathSum(i) = a[i] + a[i+k] + a[i+2k] + …

The answer is

maxCandies = max { pathSum(i) | 0 ≤ i < n } + c        // mom’s initial candies

We can compute every pathSum(i) in O(n) by scanning from right-to-left:

text
dp[i] = a[i] + (i + k < n ? dp[i + k] : 0)
best  = max(best, dp[i])

dp[i] is exactly pathSum(i) because the part after i is already known when we are walking backwards.

C++ 17 implementation

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

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

    long long n;
    if (!(cin >> n)) return 0;

    vector<long long> a(n);
    for (long long &x : a) cin >> x;

    long long k, c;
    cin >> k >> c;

    vector<long long> dp(n, 0);
    long long best = LLONG_MIN;

    for (long long i = n - 1; i >= 0; --i) {
        dp[i] = a[i];
        if (i + k < n) dp[i] += dp[i + k];
        best = max(best, dp[i]);
    }

    cout << best + c << '\n';
    return 0;
}
  • Time-complexity: O(n)
  • Memory: O(n) (can be reduced to O(k) if needed)

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