199A – Hexadecimal’s theorem

The basic approach is:


Hence, if we generate the list of all Fibonacci numbers up to (input bound) the answer is just a binary-search-away.
C++ implementation:

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

typedef unsigned long long int64;
#define N 48

int64 fib[N];
int nFib;

int64 generate() {
	fib[0] = 0;
	fib[1] = 1;
	for (int i = 2; i < N; ++i)
		fib[i] = fib[i - 1] + fib[i - 2];
	return fib[N - 1];
}

int main(void) {
	generate();
	int64 n;
	cin >> n;
	int64 * p = upper_bound(fib, fib + N, n) - 1;
	if (p - 4 >= fib)
		cout << *(p - 4) << " " << *(p - 3) << " " << *(p - 1) << endl;
	else
		cout << "0 0 " << *p << endl;
	return EXIT_SUCCESS;
}

There’s an other naive (totally valid) solution. You noticed that 0 is a Fibonacci.

#include <stdio.h>
#include <stdlib.h>

int main(void) {
	int n;
	scanf("%d", &n);
	printf("0 0 %d\n", n);
	return EXIT_SUCCESS;
}

Ironic!!!