Finalize the program

See more:

C++14

input with this test is true
10 # the number of digits in the bottom
4 2 2 2 2 2 # first digit number of degrees in this row -1
1 2 2
1 3 2
1 2 3
3 2 2 2 2
2 2 2 2
1 3 3
3 3 3 3 3
2 4 3 3
2 2 3 4

output
2 4 3 6 7 5 9 10 1 8

Here the task is to raise to a power and then compare their numbers to derive the index in the order of increasing numbers
exponentiation is from the end
2 2 3 4
is raised as 3 ^ 4 = 81
Now 2 ^ 81 = 2417851639229258349412352

if you submit such numbers, it works incorrectly
input
3
4 5 2 49 4 5
4 3 2 7 2 11
4 2 2 49 32 2

output
3 2 1 should be
my conclusion 3 1 2

#include <fstream>
#include <algorithm>
 
#include <stdio.h>
#include <math.h>
using namespace std;
 
class tower_t {
public:
    int num; 
    int height; 
    double val[11]; // содержимое
    double cache[11]; // кэш для ускорения расчета
 
    // Конструктор
    tower_t() {
        for (int i = 0; i < 11; i++) {
            val[i] = 1;
            cache[i] = 0;
        }
        height = 0;
    }
 
    // Тройной логарифм верхних 3-х уровней
     double head(int level) {
        if(cache[level] == 0) cache[level] = (log2(log2(val[level])) + log2(val[level + 1]) * val[level + 2]);
        return cache[level];
    }
 
    // Вычисление верхушки до тех пока влазит в double
    void normalize() {
        while(height > 1 && (log2(val[height - 2]) * val[height - 1]) < 50) {
            val[height - 2] = pow(val[height - 2], val[height - 1]);
            val[height - 1] = 1;
            height--;
        }
    }
 
    // Вывод для отладки
    void print() {
#ifdef _DEBUG
        printf("%2d: {", num);
        for (int i = 0; i < height; i++) {
            if (i > 0) printf(", ");
            if(val[i] < 1000000000) {
                printf("%0.0f", val[i]);
            } else {
                printf("%0.3e", val[i]);
            }
        }
        printf("}\n");
#endif
    }
};
 
// сравнение двух 
bool compare(tower_t& t1, tower_t& t2) {
    // этаж с которого сравнивать три последних уровня
    int level = ((t1.height > t2.height) ? t1.height : t2.height) - 3;
    if (level < 0) level = 0;
    if(t1.height == t2.height) { // если  одной высоты, сравниваем поэтажно
        for (int i = t1.height - 1; i >= 0; i--) {
            if (abs(t1.val[i] - t2.val[i]) > (t1.val[i] * 1e-14)) {
                if (i < level) { // верхи  совпали ниже level
                    return t1.val[i] < t2.val[i];
                }
                break;
            }
        }
    }
    return t1.head(level) < t2.head(level);
}
 
int main()
{
    // Считывание задания
    ifstream in ("input.txt");
    int cnt;
    in >> cnt;
    tower_t* towers = new tower_t[cnt];
    for (int i = 0; i < cnt; i++) {
        int len;
        in >> len;
        towers[i].num = i + 1;
        bool write = true;
        for (int j = 0; j <= len; j++) {
            int val;
            in >> val;
            if (val <= 1) write = false; // если уровень <= 1 то выше не читать
            if(write) {
                towers[i].val[j] = val;
                towers[i].height = j + 1;
            }
        }
        towers[i].print();
        towers[i].normalize();
    }
    // Сортировка
    sort(towers, towers + cnt, compare);
    // Вывод результата
    ofstream out("output.txt");
    for (int i = 0; i < cnt; i++) {
        out << towers[i].num << " ";
        towers[i].print();
    }
    out << endl;
    out.close();
    delete[] towers;
    return 0;
}

What I have tried:

number theory - Complexity class of comparison of power towers - Mathematics Stack Exchange[^]