如何创建任意数量的嵌套循环？

Question

我正在尝试创建一个程序，该程序可以在C ++语言中使用多个数字。 然后，它找到哪些运算符可以使方程式成立，并显示所有正确的可能运算。 示例：如果我放3 5 15然后输出3x5 = 15如果我放1 2 3 4 4然后输出1 + 2-3 + 4 = 4

以下代码是我编写的程序：问题是，当我想减少输入数量或增加输入数量时，我需要每次都添加/减少嵌套循环。 我想知道什么是更灵活的嵌套循环或递归方法的更有效方法。

#include <iostream>
#include <cmath>

using namespace std;


char getOperator(int);
double operate(int, double, double);

int main() {
    double a, b, c, d, e, result;
    short noOfAnswers = 0;
    cout << "Input first 5 numbers to make it equal to another 1 number.\n" <<
    "I'll find what are the operators needed to make 2 sides of the equation equal.\n";
    cin >> a >> b >> c >> d >> e >> result;
    int noOfOperators = 5;
    for (int i = 0; i <= noOfOperators; i++) {
        double firstTwo = operate(i, a, b);
        for (int j = 0; j <= noOfOperators; j++) {
            double firstThree = operate(j, firstTwo, c);
            for (int k = 0; k <= noOfOperators; k++) {
                double firstFour = operate(k, firstThree, d);
                for (int l = 0; l <= noOfOperators; l++) {
                    double firstFive = operate(l, firstFour, e);
                    if (firstFive == result) {
                        cout << ++noOfAnswers << ')' << a << getOperator(i) << b << getOperator(j) << c
                        << getOperator(k) << d << getOperator(l) << e << '=' << result << endl;
                    }
                }
            }
        }
    }
    if (noOfAnswers) cout << "I have found " << noOfAnswers << " solutions for this extremely hard problem for humanity \nin less than a second." << endl;
    else cout << "I cannot find any solutions to this problem.\n"
    <<"They're just a bunch of random numbers & That is UNSOLVABLE!" << endl;
    cout << "Do not doubt my judgment. I am always right!" << endl << "(Please note that all calculations are done from the left side first.)" << endl;
    return 0;
}

double operate(int iteration, double num1, double num2) {
    switch (iteration) {
        case 0: return num1+num2;
        case 1: return num1-num2;
        case 2: return num1*num2;
        case 3: return num1/num2;
        case 4: return pow(num1, num2);
        case 5: return fmod(num1, num2);
    }
    return 0;
}

char getOperator(int pos) {
    switch (pos) {
        case 0: return '+';
        case 1: return '-';
        case 2: return 'x';
        case 3: return '/';
        case 4: return '^';
        case 5: return '%';
    }
    return ' ';
}

Answer 1

以下内容可能会有所帮助：

// increment v where each value is a digit with maximal value maxSize
// so {0,1,2}, 3 lead to {0,2,0}
// return false on overflow.
bool increment(std::vector<int>& v, int maxSize)
{
    for (auto it = v.rbegin(); it != v.rend(); ++it) {
        ++*it;
        if (*it != maxSize) {
            return true;
        }
        *it = 0;
    }
    return false;
}

// display something like 1 + 2 * 3 = 9 // with the following meaning ((1 + 2) * 3) = 9
void display(const std::vector<double>& operands, const std::vector<int>& operators, double total)
{
    const char operators_string[] = "+-*/^%";

    std::cout << operands[0];
    for (std::size_t i = 0; i != operators.size(); ++i) {
        std::cout << " " << operators_string[operators[i]] << " " << operands[i + 1];
    }
    std::cout << " = " << total << std::endl;
}

// Compute something like {1 2 3 4}, {+ * /} as (((1 + 2) * 3) / 4)
double compute(const std::vector<double>& operands, const std::vector<int>& operators)
{
    std::function<double(double, double)> fs[] = {
        [](double a, double b) { return a + b; },
        [](double a, double b) { return a - b; },
        [](double a, double b) { return a * b; },
        [](double a, double b) { return a / b; },
        [](double a, double b) { return pow(a, b); },
        [](double a, double b) { return fmod(a, b); },
    };

    double res = operands[0];
    for (std::size_t i = 0; i != operators.size(); ++i) {
        res = fs[operators[i]](res, operands[i + 1]);
    }
    return res;
}

void display_combinaison(const std::vector<double>& operands, double total)
{
    std::vector<int> operators(operands.size() - 1);

    do {
        if (compute(operands, operators) == total) {
            display(operands, operators, total);
        }
    } while (increment(operators, 6));
}

现场例子

Answer 2

您可能要使用while（）循环，因为您不知道循环何时终止。

int main() {
    double numbers[] = {3,5,15};//consider storing the number as an array
    //the last element is the result
    double result;
    int arr_len = sizeof(numbers)/sizeof(double);
    int i,j;

    while(1)
    {
          j = 0;
          while(j++ < 5)//over all operators
          {i = 0;
              result = numbers[0];//start with first element
              while(i < arrlen - 2)//over all numbers, exclude the result
              {
                 result = operate(j, result, numbers[++i]);
                 //something like this...this does not work correctly
                 //it might give you a hint in the right direction
                 if(result == numbers[arr_len - 1])//compare to last element
                     return 0;
              }  
          }
    }
    return 0;
}