logoChat01.ai

Gibe in cpp

تم الإنشاء في: ١٢ يوليو ٢٠٢٥

تمت الإجابة باستخدام O3 بواسطة Chat01

سؤال

17523166198083311739035655886738.jpg17523166198083311739035655886738.jpg
17523166527986043425490551793870.jpg17523166527986043425490551793870.jpg
17523166643354484832538178048758.jpg17523166643354484832538178048758.jpg

Gibe in cpp

الفكرة

Transcribing problem statement

I notice the user posted three images with a code outline for "leastButtonPress," but the code appears incomplete and there’s a misspelling ("Boutton" instead of "Button"). They're asking for a C++ solution to the given problem, which involves navigating stations with a number pad and specific digits and buttons. I'll read the images for the problem details. The first image mentions the Large Hadron Collider stations and involves figuring out the fastest way to reach a station using working digits on the number pad, with N stations in total.

Clarifying problem rules

The problem involves navigating stations using functioning digits on a number pad. Buttons E, F, and B are always working and help with station movement. The goal is to find the shortest sequence of button presses, considering valid station numbers and digit limitations. The approach is to use BFS or Dijkstra's algorithm to evaluate all reachable stations from the starting position, calculating the minimum presses based on allowed moves (digit input and Forward/Backward). I need to carefully account for all valid station numbers and transitions for a solution.

Evaluating shortest path algorithm

The challenge is to calculate the minimum number of button presses to go from a starting station to a destination. The simpler approach involves calculating the absolute difference using Forward/Backward moves, but we also have to account for typing stations using available digits and the Enter button. Multiple Enter actions could be useful if they reduce the total cost, but it seems unlikely. To compute this efficiently, I'll use Dijkstra’s algorithm to assess each station’s cost, considering typable stations and the cost of moving between them.

Simplifying approach with minimal Enter

It looks like I don’t need to worry too much about multiple Enter actions after the first. The cost of typing any station via Enter is independent of where I start, so I can treat them separately. Plus, F/B moves are just about the distance, with no additional complexity. So, I’ll focus on calculating the cost for each typable station (digit) plus the distance from there to the destination. Multiple Enter presses don’t seem worth considering—just one would do.

Thought for 1m 22s

الإجابة

Here’s a straightforward O(N × log N) solution that scans every station once, checks whether its digits are type-able with the working keys, and keeps the best answer.

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

/* ---------- helpers ---------- */
int ringDist(int a, int b, int n) {                // shortest F/B distance on a ring
    int forward  = (b - a + n) % n;                // move with F
    int backward = (a - b + n) % n;                // move with B
    return min(forward, backward);
}

bool canType(int x, const vector<bool>& ok) {      // all digits work?
    if (x == 0) return ok[0];
    while (x) {
        if (!ok[x % 10]) return false;
        x /= 10;
    }
    return true;
}

int digitCount(int x) {                            // #digits in x (x ≥ 0)
    if (x == 0) return 1;
    int c = 0;
    while (x) { ++c; x /= 10; }
    return c;
}

/* ---------- core ---------- */
int leastBouttonPress(int n, int d, int digits[], int pres, int dest) {
    vector<bool> ok(10, false);
    for (int i = 0; i < d; ++i) ok[digits[i]] = true;

    int best = ringDist(pres, dest, n);            // only F/B keys

    for (int s = 1; s <= n; ++s) {                 // try every type-able station
        if (!canType(s, ok)) continue;
        int cost = digitCount(s) + 1               // press digits of s, then Enter
                 + ringDist(s, dest, n);           // finish with F/B to dest
        best = min(best, cost);
    }
    return best;
}

/* ---------- driver ---------- */
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, d, pres, dest;
    if (!(cin >> n)) return 0;
    cin >> d;
    vector<int> digs(d);
    for (int &x : digs) cin >> x;
    cin >> pres >> dest;

    cout << leastBouttonPress(n, d, digs.data(), pres, dest);
    return 0;
}

Why it works

  • ringDist handles the circular track (forward vs. backward moves).
  • canType checks if a station’s digits are all among the functioning keys.
  • For every station you could teleport to (type + Enter), the total cost is
    digits of s + 1 (Enter) + ringDist(s, dest).
  • The minimum of all those possibilities, plus the direct F/B walk, is the answer.

With N ≤ 10 000, this runs easily within limits.

مشاركة هذا السؤال والجواب