Program about Fibonacci numbers - c++

Question

I am obliged to write program that will show minimal quantity of Fibonacci numbers which sum is equal to k number which user write into program.
So for example user writes 1070
And the program will show 987 89 -5 -1 (because all of these numbers are Fibonacci numbers (you can add or substract to get desired sum).
I have problem to think about solution to this problem.
I wrote code to get all Fibonacci numbers from 0 to 46.
#include <iostream>
unsigned long long Fibbo[47];
void preapare()
{
Fibbo[0] = 0;
Fibbo[1] = 1;
i = 2;
while (i<47)
{
Fibbo[i] = Fibbo[i - 1] + Fibbo[i - 2];
i++;
}
}
int main()
{
preapare();
int k=0;
std::cin >> k;
}
I will be glad for any help. You can use Fibonacci Number as many times as you will. For example you can do 1+1+1+1+1

Answer 1

#include <iostream>
#include <vector>
#include <stdlib.h>
using namespace std;
void prepare( vector<int> & fibos ) { ... } // along the lines we see in OPs code
int findClosest( int n, vector<int> & fibos ){ // return Fibonacci number closest to n
int fi = 0;
for( std::vector<int>::iterator it = fibos.begin() ; it != fibos.end(); ++it){
if( abs(*it - n) < abs(fi - n) ){
fi = *it;
}
}
return fi;
}
// The algorithm:
// lookup closest Fi, add "signed" to terms
// reduce n, adjust sign and call recursively
void sum( int n, vector<int> & fibos, vector<int> & terms, int sign = 1 ){
if( n == 0 ) return;
int fi = findClosest( n, fibos );
terms.push_back( sign*fi );
sum( abs(n - fi), fibos, terms, n - fi > 0 ? sign : -sign );
}
int main() {
vector<int> fibos;
prepare( fibos );
vector<int> terms;
int n = 1070;
sum( n, fibos, terms );
cout << n << " =";
for( std::vector<int>::iterator it = terms.begin() ; it != terms.end(); ++it){
cout << " " << *it;
}
cout << endl;
return 0;
}

Answer 2

For user input:
Before the while loop, you need to get the input, save it to a variable, and use that in your while condition. Something like this:
#include <iostream>
...
int k_val = 0;
std::cout << "Enter the value for k: ";
std::cin >> k_val;
...

Answer 3

Your prepare() fixed:
unsigned long long* preapare(const size_t n) {
unsigned long long* Fibbo = new unsigned long long[n];
Fibbo[0] = 0;
Fibbo[1] = 1;
for(size_t i = 2; i < n; ++i) {
Fibbo[i] = Fibbo[i - 1] + Fibbo[i - 2];
}
return Fibbo;
}
then you'll have to memory manage the returned array;

Answer 4

I've done this, the code will solve the grater fibonacci numbers to be add so the result will be the specified number:
fiboMax return a sequence with the fibonacci number less to the input number.
getiboSec return a sequence with the bigger fibonacci numbers that sum the input number.
'''
Created on 15/01/2014
'''
def fiboMax(num):
ret = [0,1]
a = 1
b = 2
while b < num:
ret.append(b)
tmp = a + b
a = b
b = tmp
return ret
def getFiboSec(num):
fiboSec = []
fiboNumbers = fiboMax(num)
i = len(fiboNumbers) - 1
while num > 0 and i > 0:
fiboNum = fiboNumbers[i]
while(fiboNum <= num):
fiboSec.append(fiboNum)
num -= fiboNum
i -= 1
return fiboSec
if __name__ == '__main__':
num = 20
print(fiboMax(num))
print(getFiboSec(num))
output:
figonnaci number to 20 = [0, 1, 2, 3, 5, 8, 13]
fibonacci number that sum 20 = [13, 5, 2]
It doesnt work with negative numbers, anyway its solving the problem but not the efficient way.

Answer 5

You are accessing the indexes -2 and -1 in your loop. Your variable i should start at 2. You also try to add values to pointers. None of your values in Fibbo[] are set.