Онлайн компилятор C

#include <stdio.h> void print_times(int m, int times) { if (times == 0) { return; } printf("%d ", m); print_times(m, times - 1); } int current = 1; void print_sequence(int n) { if (n <= 0) { return; } if (n >= current) { print_times(current, current); print_sequence(n - current); current++; } else { print_times(current, n); } } int main() { int x; scanf("%d", &x); print_sequence(x); return 0; }
This code prints a sequence of numbers where each number appears a certain number of times. The function `print_times` prints the number `m` exactly `times` times. The function `print_sequence` uses a global variable `current` to decide which number to print next and how many times. It tries to print `current` repeated `current` times, then subtracts that count from `n` and increments `current`. If `n` is smaller than `current`, it prints `current` only `n` times.

- The global variable `current` is modified inside `print_sequence`, but its value persists across recursive calls. Check if the logic for updating `current` and subtracting from `n` correctly handles the case when `n` is not exactly a sum of consecutive numbers like 1 + 2 + 3 + ... . For example, if `x` is 5, what sequence do you expect, and what does the code produce?
- The condition `if (n >= current)` uses the current value of `current` before it is incremented. Trace through a small example like `x = 2` step by step to see if the recursion and the increment of `current` work as intended.