如何找到添加兩個變量的所有可能組合，每個變量都附加到一個乘數，總和為給定的數字（cin）？

Question

在我的情況下，一輛卡車的容量為 30，而一輛面包車的容量為 10。我需要找到運輸給定數量的貨物所需的貨車/卡車的數量，比如 100。我需要找到所有可能的組合卡車 + 廂式貨車加起來為 100。

基本的數學計算是：(30*lorrycount) + (10*vancount) = n，其中 n 是貨物數量。

輸出示例

運輸貨物：100

貨車數量：0 3 2 1

貨車數量：10 1 4 7

例如，第二個組合是 3 輛卡車，1 輛面包車。 考慮到卡車的容量 = 30，貨車容量 = 10，(30*3)+(10*1) = 100 = n。

目前，我們只有這段代碼，它可以在不考慮上面給出的公式的情況下，從字面上查找加起來為給定數字 n 的所有數字組合。

#include <iostream>
#include <vector>
using namespace std;

void findCombinationsUtil(int arr[], int index,
    int num, int reducedNum)
{
    int lorry_capacity = 30;
    int van_capacity = 10;
    // Base condition 
    if (reducedNum < 0)
        return;

    // If combination is found, print it 
    if (reducedNum == 0)
    {
        for (int i = 0; i < index; i++)
            cout << arr[i] << " ";
        cout << endl;
        return;
    }

    // Find the previous number stored in arr[] 
    // It helps in maintaining increasing order 
    int prev = (index == 0) ? 1 : arr[index - 1];

    // note loop starts from previous number 
    // i.e. at array location index - 1 
    for (int k = prev; k <= num; k++)
    {
        // next element of array is k 
        arr[index] = k;

        // call recursively with reduced number 
        findCombinationsUtil(arr, index + 1, num,
            reducedNum - k);
    }
}


void findCombinations(int n)
{
    // array to store the combinations 
    // It can contain max n elements 
    std::vector<int> arr(n); // allocate  n elements

    //find all combinations 
    findCombinationsUtil(&*arr.begin(), 0, n, n);
}
int main()
{
    int n;
    cout << "Enter the amount of cargo you want to transport: ";
    cin >> n;
    cout << endl;
    //const int n = 10;
    findCombinations(n);

    return 0;
}

如果您有任何解決方案，請告訴我，謝謝。

Answer 1

我們將創建一個遞歸函數，從左到右遍歷全局capacities數組，並嘗試將貨物裝載到各種車輛類型中。 我們跟蹤我們仍然需要加載多少並將其傳遞給任何遞歸調用。 如果到達數組的末尾，則僅當剩余貨物為零時才生成解決方案。

std::vector<int> capacities = { 30, 10 };
using Solution = std::vector<int>;
using Solutions = std::vector<Solution>;

void tryLoad(int remaining_cargo, int vehicle_index, Solution so_far, std::back_insert_iterator<Solutions>& solutions) {
    if (vehicle_index == capacities.size()) {
        if (remaining_cargo == 0) // we have a solution
            *solutions++ = so_far;
        return;
    }
    int capacity = capacities[vehicle_index];
    for (int vehicles = 0; vehicles <= remaining_cargo / capacity; vehicles++) {
        Solution new_solution = so_far;
        new_solution.push_back(vehicles);
        tryLoad(remaining_cargo - vehicles * capacity, vehicle_index + 1, new_solution, solutions);
    }
}

如下調用它應該在all_solutions產生所需的輸出：

Solutions all_solutions;
auto inserter = std::back_inserter(all_solutions)
tryLoad(100, 0, Solution{}, inserter);

Answer 2

尋找所有可能組合的迭代方法

#include <iostream>
#include <vector> 

int main()
{
    int cw = 100;
    int lw = 30, vw = 10;

    int maxl = cw/lw;  // maximum no. of lorries that can be there
    std::vector<std::pair<int,int>> solutions;    

    // for the inclusive range of 0 to maxl, find the corresponding no. of vans for each variant of no of lorries
    for(int l = 0; l<= maxl; ++l){
        bool is_integer = (cw - l*lw)%vw == 0; // only if this is true, then there is an integer which satisfies for given l

        if(is_integer){
            int v = (cw-l*lw)/vw; // no of vans
            solutions.push_back(std::make_pair(l,v));
        }
    }

    for( auto& solution : solutions){
        std::cout<<solution.first<<" lorries and "<< solution.second<<" vans" <<std::endl;
    }
    return 0;
}