问Ackermann函数的迭代实现

EN

#include <stack>
#include <iostream>
#include <tuple>
#include <vector>

int Ackermann(int m, int n) {
    std::stack<int> s;
    s.push(m);
    while (!s.empty()) {
        m = s.top();
        s.pop();
        if (m == 0 || n == 0)
            n += m + 1;
        else {
            s.push(m - 1);
            s.push(m);
            n--;
        }
    }
    return n;
}

int main() {
    std::vector<std::tuple<int, int, int>> tests{
        { 0, 0, 1},
        { 1, 0, 2},
        { 1, 1, 3},
        { 2, 1, 5}
    };

    for (auto const &test : tests) {
        using std::get;

        auto result = Ackermann(get<0>(test), get<1>(test));
        if (result == get<2>(test)) {
            std::cout << "Ackermann(" 
                       << get<0>(test) << ", " 
                       << get<1>(test) << ") == " 
                       << result << ": passed\n";
        }
        else {
            std::cerr << "Error: Ackermann(" 
                      << get<0>(test) << ", " 
                      << get<1>(test) 
                      << ": expected" << get<2>(test) 
                      << ", but got: " << result << "\n";
        }
    }
}