Simulating recursion (1) P62390
In C++, consider this program (whose inclusions have been removed):
void work(int n) {
if (n > 0) {
cout << ' ' << n;
work(n - 1);
work(n - 1);
}
}
int main() {
int n;
while (cin >> n) {
work(n);
cout << endl;
}
}
In Python, consider this program:
from yogi import tokens
def work(n: int) -> None:
if n > 0:
print(' ', n, end='')
work(n - 1)
work(n - 1)
def main() -> None:
for n in tokens(int):
work(n)
print()
main()
Take a look at the sample input and sample output to see what this program prints for every given number.
Without modifying main(), reimplement the procedure work(n) with no calls at all, recursive or not, so that the output of the program does not change.
Input
Input consists of several strictly positive natural numbers.
Output
For every number, print a line identical to the one written by the program above.
Observation
To solve this exercise, the only containers that you should use are stacks.
Public test cases
Input
1 2 3 4
Output
1 2 1 1 3 2 1 1 2 1 1 4 3 2 1 1 2 1 1 3 2 1 1 2 1 1
Information
- Author
- Salvador Roura
- Language
- English
- Translator
- Carlos Molina
- Original language
- Catalan
- Other languages
- Catalan
- Official solutions
- C++ Python
- User solutions
- C++ Python