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A system contains n files, where the size of the t...

Erstellt am: 7. November 2025

Beantwortet mit GPT-5 Thinking von Chat01

Frage

A system contains n files, where the size of the th file is fileSizell. To protect against hacking, each file must have a minimum size of minSize(i).

You can transfer content between files to protect them all, but the total number of bytes must remain the same. Your task is to find the minimum number of files whose sizes must be changed to ensure all files meet their minimum size requirements. If it is not possible to protect all files, return -1.

Example
n = 6
fileSize [4, 1, 5, 2, 3)

minSize [3, 2, 2, 1.4]

One solution is to transfer 2 bytes from file 3 (Index 2). distributing 1 byte each to file 2 (index 1) and file 5 (index 4).

After the transfer:

fileSize [4, 2, 3.2,41

Now for each I, fileSize(i) z minSize(), so all files are protected. The minimum number of files whose sizes must be changed is 3 (files at Indices 1, 2, and 4).

Function Description

Complete the function getMinimumChanges in the editor with the following parameters:

int fileSize(n): the initial sizes of each file

file int minSize(n): the minimum required sizes to protect each

Returns

Int: the minimum number of files whose sizes need to be changed to protect all the files
Constraints

Isns2 105

ST

1s fileSizelil, minSiamfil s 109

Input Format For Custom Testing

Sample Case O

Sample Input For Custom Testing

11

STODI

14

FUNCTION

The functions

15

filesize[] size, n.3

filesize [2, 5, 1]

minsize[] size, n=3

minsize [4, 4, 4]

Sample Output

-1

Explanation

There is no way to protect all the files.

Sample Case 1

Sample Input For Custom Testing

STDIN

5

2

3

1

5

FUNCTION

filesize[] size, n=5

filesize [2, 3, 1, 5, 9]

minsize[] size, n-5

minsize [1, 5, 6, 1, 5)

Sample Output

4

Explanation

Considering 1-based indexing, the size of files 2, 3, 4, and 5 can be changed to fileSize = 12, 5, 6, 1, 61. Now, for each fileSizelil z minSizelil, and thus all files are protected.

Give me c++ 20 program of above problem according to following function

]

Sample Output

-1

Explanation

There is no way to protect all the files.

▼Sample Case 1

Sample Input For Custom Testing

STDIN

FUNCTION

filesizell size, n=5

2

Language C++20

9

1 > #include <bits/stdc++.h>

S2

Complete the 'getMinimumChanges' function below.

13

2

The function is expected to return an INTEGER.

16

3

The function accepts following parameters:

  1. INTEGER ARRAY filesize

  2. INTEGER ARRAY minsize
    int getMinimumChanges (vector<int> filesize, vectorcist ninsize)

int main()...

Gedanke

Clarifying the file size problem

I need to write a C++20 program that ensures each file size matches or exceeds a minimum size requirement. The total size must stay the same, and I’m looking for the minimum number of files where their sizes need to be modified. If it's not possible, the program should return -1. I’ll check regarding the sum of file sizes.

Optimizing file size adjustments

To minimize the number of file changes, we should select files with surplus (fileSize[i] - minSize[i]) to support those with deficits. The goal is to reduce the number of donor files used, ensuring each donor is only adjusted once. The minimum changes are at least the number of deficit files, and we'll try to minimize the donors needed to cover those deficits.

Finalizing the approach for file adjustments

The minimal number of files to change is at least the number of deficit files. If there are no deficits, no changes are needed. Donors are only essential when deficits exist, and I'd need to decrease some donors to cover the deficits. The smallest number of donors should cover the total deficit, while files with exactly equal sizes to minSize don't need adjustment.

Finalizing the algorithm

Okay, I think I’ve got it:

  1. Compute the deficit as need[i] = max(0, minSize[i] - fileSize[i]).
  2. Sum these to get total_deficit and count files needing changes, D.
  3. Compute surpluses for files, sum them to get total_surplus.
  4. If surplus is less than deficit, return -1.
  5. If no deficit, return 0.
  6. Sort surpluses and pick files with the largest surpluses until the deficit is covered.
  7. The final answer is the number of deficit files, D, plus the number of donors, K.

Thought for 39s

Antwort

Here’s a clean C++20 solution that computes the minimum number of files that must change. It’s based on counting all deficit files (those below their min) plus the fewest donors needed (take from the largest surpluses until the total deficit is covered). If the total deficits can’t be covered by surpluses, return -1.

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

// Returns the minimum number of files whose sizes must be changed,
// or -1 if it's impossible to meet all minimums.
int getMinimumChanges(vector<int> filesize, vector<int> minsize) {
    const int n = (int)filesize.size();

    long long totalDeficit = 0;
    int deficitCount = 0;
    vector<long long> surplus; 
    surplus.reserve(n);

    for (int i = 0; i < n; ++i) {
        long long fs = filesize[i];
        long long ms = minsize[i];
        if (fs < ms) {
            totalDeficit += (ms - fs);
            ++deficitCount;              // these files must be increased => must change
        } else if (fs > ms) {
            surplus.push_back(fs - ms);  // potential donors
        }
        // if fs == ms: neither deficit nor surplus; we prefer to leave it unchanged
    }

    // If nothing is below minimum, no changes are needed.
    if (totalDeficit == 0) return 0;

    // Use the fewest donors: take largest surpluses first until covering totalDeficit.
    sort(surplus.begin(), surplus.end(), greater<long long>());

    long long taken = 0;
    int donorCount = 0;
    for (long long s : surplus) {
        taken += s;
        ++donorCount;
        if (taken >= totalDeficit) break;
    }

    if (taken < totalDeficit) return -1; // Not enough total bytes to cover deficits

    // Minimum files changed = number of deficit files + number of donors used
    return deficitCount + donorCount;
}

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

    // Input format:
    // n
    // n integers: filesize
    // n integers: minsize
    int n;
    if (!(cin >> n)) return 0;

    vector<int> filesize(n), minsize(n);
    for (int i = 0; i < n; ++i) cin >> filesize[i];
    for (int i = 0; i < n; ++i) cin >> minsize[i];

    cout << getMinimumChanges(filesize, minsize) << "\n";
    return 0;
}

How it works (brief)

  • Count deficits and total bytes needed.
  • Gather all positive surpluses.
  • If no deficits → answer 0.
  • Otherwise, sort surpluses descending and pick the smallest number of donor files whose sum covers the total deficit.
  • Answer = (number of deficit files) + (donors chosen). If surpluses can’t cover deficits, return -1.

Frage

A company buys fragile components that are shipped in costly containers. The seller offers a promotion: customers receive a free container of components if they return a specific number of empty containers.

Calculate the maximum number of containers a customer can obtain given:

  1. A fored budget

  2. The cost per container

  3. The number of empty containers required for a free container

Each container is assumed to be emptied and returned for credit immediately.

Example

0-4

cost-1

m-2

Start with a budget n = 4 units of currency to buy containers at cost = 1 unit each. The number of empty containers to retum for a free full container is m 2. First, buy 4 containers with the entire budget. Turn in thosa 4 containers for 2 more containers. Tum in those 2 containers for one mors. At this point, there are not enough funds or containers to receive another. In total, 4+2+1-7 containers were obtained.

Function Description

Complete the function maximum Containers in the editor with the following parameter(s):

string scenarios[n]: Each string contains three space-separated integers, starting budget, cost, and the number of containers to return for a full container.
Print

For each test case, the function must print an integer: the maximum number of containers the manufacturer can obtain, each on a new line. No return value is expected.

Constraints

Isns103

2 s starting budget s 105

1ss cost of a container in money is starting budget

2 s number of empty containers to retum s starting budget

Input Format for Custom Testing

Input from stdin will be processed as follows and passed to the function.

The first line contains an integer n, the size of scenariod or the number of test cases.

Each of the next n lines contains three space-separated integers, original budget, cost in dollars, and cost in empty containers. These are read as a single string by the code stub.

Sample Case O

Sample Input O

STDIN

3

Function

10 25

size of scenarios[] n = 3

'6 2 2']

scenarios ['10 2 5', '12 44',

12 4 4

622

Sample Output O

6

Language

C++20

#include bits/sto

S2

2

Complete the 'ww

13

14

The function ac

15

17

18

19

20> int main()

Explanation O

There are 3 test cases:

  1. Spend 10 units of currency on 6 containers at 2 units sach. Turn in 6 conteiners for another one for a total of 6.

  2. Spend 12 units on 3 containers at 4 units. There are not enough containers to turn in, so the total is 3.

  3. Spend 6 units on 3 containers at 2 units. Tum in 2 of the containers for a 4th container leaving 1 old and I new conteiner. Turn those 2 in for 1 more container making the total 6.

Sample Case 1

Sample Input 1

STDIN

2

842

Function

scenarios[] size n = 2

723

scenarios ['8 42', '723']

Sample Output 1

10

11

Explanation 1

There are 2 test cases:

  1. Spend 8 units of currency on 2 containers at 4 units each. Turn in 2 containers for another one.
  2. Spend 6 units on 3 containers at 2 units. Tum in 3 containers for another one. 1 unit of currency is left over.

Give c++ 20 program of above problem according to following function

#include <bits/stdc++.h>...

Complete the 'maximumContainers' function below.

The function accepts STRING ARRAY scenarios as parameter.

void maximumContainers (vector<string> scenarios)I
int main()...

Gedanke

Designing C++20 for maxContainers

I need to write a C++20 program for the maximumContainers function. It's based on the "chocolate feast" problem, considering a starting budget, container cost, and exchange rates. I'll parse input for multiple test cases and ensure the program handles large numbers (up to 1e5) with appropriate safeguards using long long. I'll provide clear parsing and output structure.

Implementing the algorithm for maxContainers

I need to implement the algorithm considering the initial budget, cost per container, and exchange rate for empties. After calculating the initial purchases and setting up the exchange trade while checking for remaining empties, I'll handle edge cases carefully, like when the cost exceeds the budget. I'll ensure proper input/output structure, printing results correctly per user request.

Building robust C++ code

I'll make the code robust by using istringstream to parse each scenario and handle stray spaces carefully. It’s unnecessary to validate negative numbers, but I’ll ensure m >= 2 and cost > 0 to avoid infinite loops. I’ll compute purchases, total containers, and handle the exchange rate properly. Then, I’ll accumulate results into a vector for each test case and ensure correct input reading.

Finalizing C++20 implementation

For this task, I'll start by reading the given number of test cases, handle potential blank lines with a loop, and store the input in a vector<string>. The function maximumContainers will process each scenario and print results directly. I'll ensure we're using C++20 features and avoid async behavior. The explanation will stay concise, just focusing on the code.

Thought for 23s

Antwort

Here’s a straightforward C++20 implementation that matches the requested function signature and I/O. It parses each scenario line, simulates buying and exchanging containers, and prints the maximum obtainable for each case.

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

/*
 * Complete the 'maximumContainers' function below.
 *
 * The function accepts STRING ARRAY scenarios as parameter.
 * It prints one integer per line (no return value).
 */
void maximumContainers(vector<string> scenarios) {
    ios::fmtflags f(cout.flags()); // preserve formatting, just in case

    for (const string& s : scenarios) {
        long long budget = 0, cost = 0, m = 0;

        // Parse "budget cost m" from the scenario string.
        {
            istringstream iss(s);
            iss >> budget >> cost >> m;
        }

        // Guard against invalid cost or exchange (per constraints m >= 2, cost >= 1).
        if (cost <= 0 || m <= 1 || budget < cost) {
            cout << (budget >= cost && m > 1 ? budget / cost : 0) << "\n";
            continue;
        }

        long long bought = budget / cost;  // initial containers purchased
        long long total = bought;          // total containers obtained
        long long empties = bought;        // empty containers available to exchange

        // Exchange empties for full containers as long as possible.
        while (empties >= m) {
            long long trade = empties / m;        // free full containers obtained
            total += trade;
            empties = trade + (empties % m);      // new empties after using/freeing
        }

        cout << total << "\n";
    }

    cout.flags(f);
}

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

    int n;
    if (!(cin >> n)) return 0;
    string dummy;
    getline(cin, dummy); // consume end of line after n

    vector<string> scenarios;
    scenarios.reserve(n);

    for (int i = 0; i < n; ++i) {
        string line;
        // Read non-empty line (handle possible blank lines in custom input)
        do {
            if (!std::getline(cin, line)) line.clear();
        } while (line.empty() && cin);
        scenarios.push_back(line);
    }

    maximumContainers(scenarios);
    return 0;
}

How it works (quickly):

  • Buy as many as budget / cost.
  • Repeatedly trade m empties for 1 full container, add to total, and keep the new empty.
  • Loop runs in ~log steps (fast), prints one answer per scenario line.

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