斐波那契函数的问题。 C ++

Question

Should return the n place of the array. But instead of the value I'm only getting 0.

int fibonacci(int n)
{
    int f[100];
    f[0] = 0;
    f[1] = 1;

    for (int i=2; i<n; i++)
    {
        f[i] = f[i-2] + f[i-1];
    }

    return f[n];
}

int main()
{
    cout << fibonacci(3);
    return 0;
}

New CODE:

New problem its returning one number further then it should. For example if 'n==7' its returning '13' not '8' like it should.

int fibonacci(int n)
{
    int f[100] = { 0, 1 };

    for (int i=2; i<=n; i++)
    {
        f[i] = f[i-2] + f[i-1];
    }

    return f[n-1];
}

int main()
{
    cout << fibonacci(7);
    return 0;
}

Answer 1

Your loop termination condition is wrong. Since this is homework perhaps you can work out why.

Answer 2

well, you never set f[n] , you only go up to i < n , that is, i == n-1 . try returning f[n-1]

EDIT: as Chris Lutz pointed out my answer is no good as it would give an invalid result if you called fibonacci(0)

Like many have answered already, the best solution is to loop until i <= n
Unless, of course, you want fibonacci(3) to return the 3rd element in the fibonacci sequence and not the 4th, in which case fibonacci(0) wouldn't really make sense, and the right return value would be f[n-1] ... still the n==0 case should be handled somehow, as should the n<0 and the n>100 cases.

you can return f[n-1] as long as you check for the right boundaries:

int fibonacci(int n)
{
    int f[100] = { 0, 1 };

    if ((n <= 0) || (n > 100))
        return -1;//return some invalid number to tell the caller that he used bad input

    for (int i=2; i < n; i++) // you can use i < n here
    {
        f[i] = f[i-2] + f[i-1];
    }

    return f[n-1];
}

Answer 3

You forgot to initialize the n-th value of the array. You return f[n] but only initialize up to n-1.

Answer 4

The trouble is that you test for i < n (where n == 3 in your example call), but you return f[3] which has not been set to anything. You are 'lucky' that you're getting zeroes rather than random garbage.

Change the ' < ' to ' <= '.

---

Working Code #1

Retaining the full size array.

#include <iostream>
using namespace std;

static int fibonacci(int n)
{
    int f[100] = { 0, 1 };

    if (n < 0 || n > 100)
        return -1;
    else if (n < 2)
        return f[n];

    for (int i = 2; i <= n; i++)
    {
        f[i] = f[i-2] + f[i-1];
        //cout << "f[" << i << "] = " << f[i] << endl;
    }

    return f[n];
}

int main()
{
    for (int i = 0; i < 8; i++)
        cout << "fib(" << i << ") = " << fibonacci(i) << endl;
    return 0;
}

Sample Output #1

fib(0) = 0
fib(1) = 1
fib(2) = 1
fib(3) = 2
fib(4) = 3
fib(5) = 5
fib(6) = 8
fib(7) = 13

Working Code #2

This uses an array of size 3, at the cost of a lot of modulo operations:

#include <iostream>
using namespace std;

static int fibonacci(int n)
{
    int f[3] = { 0, 1, 0 };

    if (n < 0 || n > 100)
        return -1;
    else if (n < 2)
        return f[n];

    for (int i = 2; i <= n; i++)
    {
        f[i%3] = f[(i-2)%3] + f[(i-1)%3];
        //cout << "f[" << i << "] = " << f[i%3] << endl;
    }

    return f[n%3];
}

int main()
{
    for (int i = 0; i < 8; i++)
        cout << "fib(" << i << ") = " << fibonacci(i) << endl;
    return 0;
}

It produces the same output - so there is no point in repeating it.

Working Code #3

Avoiding arrays and modulo operations:

#include <iostream>
using namespace std;

static int fibonacci(int n)
{
    int f0 = 0;
    int f1 = 1;

    if (n < 0 || n > 46)
        return -1;
    else if (n == 0)
        return f0;
    else if (n == 1)
        return f1;

    int fn;
    for (int i = 2; i <= n; i++)
    {
        int fn = f0 + f1;
        f0 = f1;
        f1 = fn;
        //cout << "f[" << i << "] = " << fn << endl;
    }

    return f1;
}

int main()
{
    for (int i = -2; i < 50; i++)
        cout << "fib(" << i << ") = " << fibonacci(i) << endl;
    return 0;
}

The limit 46 is empirically determined as correct for 32-bit signed integers.

Example Output #3

fib(-2) = -1
fib(-1) = -1
fib(0) = 0
fib(1) = 1
fib(2) = 1
fib(3) = 2
fib(4) = 3
fib(5) = 5
fib(6) = 8
fib(7) = 13
fib(8) = 21
fib(9) = 34
fib(10) = 55
fib(11) = 89
fib(12) = 144
fib(13) = 233
fib(14) = 377
fib(15) = 610
fib(16) = 987
fib(17) = 1597
fib(18) = 2584
fib(19) = 4181
fib(20) = 6765
fib(21) = 10946
fib(22) = 17711
fib(23) = 28657
fib(24) = 46368
fib(25) = 75025
fib(26) = 121393
fib(27) = 196418
fib(28) = 317811
fib(29) = 514229
fib(30) = 832040
fib(31) = 1346269
fib(32) = 2178309
fib(33) = 3524578
fib(34) = 5702887
fib(35) = 9227465
fib(36) = 14930352
fib(37) = 24157817
fib(38) = 39088169
fib(39) = 63245986
fib(40) = 102334155
fib(41) = 165580141
fib(42) = 267914296
fib(43) = 433494437
fib(44) = 701408733
fib(45) = 1134903170
fib(46) = 1836311903
fib(47) = -1
fib(48) = -1
fib(49) = -1

Answer 5

n is the index that is never reached in your version. You just need to replace the < with <= in your for loop conditional. (You never assigned f[n] because n was never reached by the loop and so you got back a default value.)

int fibonacci(int n)
{
    int f[100];
    f[0] = 0;
    f[1] = 1;

    for (int i=2; i<=n; i++)
    {
        f[i] = f[i-2] + f[i-1];
    }

    return f[n];
}

int main()
{
    cout << fibonacci(3);
    return 0;
}

And you don't need an array to perform the fib sequence by the way. Just use two variables and reassign them in the loop. Something like this:

int a = 0;
int b = 1;

for (int i=2; i<=n; i++)
{
    b = a + b;
    a = b;
}

return b;

Answer 6

With the call fibonacci(3) , your for loop (inside the fibonacci function) goes until i < 3 ...

It means that the last assigment is f[2] . Not f[3] as expected (which is the value you return).

Answer 7

Don't you mean return f[n-1];

I guess your compiler has set the array f[100] to 0?

Looks like the other guy has the right answer....