CODEWITHSUNDEEP
c++fibonacci
Juan Sierra
Juan Sierra

Reputation: 75

Output a 15 digit number

I have a program that is suppose to find the sum of all the numbers between 1 and 75 in the fibonacci sequence that are divisible by three and add them together. I got the program working properly the only problem I am having is being able to display such a large number. I am told that the answer should be 15 digits. I have tried long long, long double, unsigned long long int and none of those produce the right output (they produce a negative number).

code:

 long fibNum(int kth, int nth);

int main()
{
    int kTerm;
    int nTerm;

    kTerm = 76;
    nTerm = 3;

    std::cout << fibNum(kTerm, nTerm) << std::endl;

    system("pause");
    return 0;
}

long fibNum(int kth, int nth)
{
    int term[100];
    long firstTerm;
    long secondTerm;
    long exactValue;

    int i;

    term[1] = 1;
    term[2] = 1;
    exactValue = 0;

    do
    {
        firstTerm = term[nth - 1];
        secondTerm = term[nth - 2];
        term[nth] = (firstTerm + secondTerm);

        nth++;
    }
    while(nth < kth);

    for(i = 1; i < kth; i++)
    {
        if(term[i] % 3 == 0)
        {
            term[i] = term[i];
        }
        else
            term[i] = 0;

        exactValue = term[i] + exactValue;
    }

    return exactValue;

I found out that the problem has to do with the array. The array cannot store the 47th term which is 10 digits. Now I have no idea what to do

Upvotes: 1

Views: 2856

Answers (3)

RalfFriedl
RalfFriedl

Reputation: 1130

The array cannot store the 47th term which is 10 digits.

This indicates that your architecture has a type long with just 32 bits. That is common for 32-bit architecture. 32 bits cover 9 digit numbers and low 10-digit numbers, to be precise 2.147.483.647 for long and 4.294.967.295 for unsigned long.

Just change your long types to long long or unsigned long long, including the return type of fibNum. That would easily cover 18 digits.

Upvotes: 0

Kushal Dev
Kushal Dev

Reputation: 1

What you can do is take char s[15]and int i=14,k, and then go for while loop till sum!=0 Under while body

k=n%10;
s[i]=k+48;
n=n/10;
i--;

Upvotes: 0

Keith Thompson
Keith Thompson

Reputation: 263197

Type long long is guaranteed to be at least 64 bits (and is exactly 64 bits on every implementation I've seen). Its maximum value, LLONG_MAX is at least 263-1, or 9223372036854775807, which is 19 decimal digits -- so long longis more than big enough to represent 15-digit numbers.

Just use type long long consistently. In your code, you have one variable of type long double, which has far more range than long long but may have less precision (which could make it impossible to determine whether a given number is a multiple of 3.)

You could also use unsigned long long, whose upper bound is at least 264-1, but either long long or unsigned long long should be more than wide enough for your purposes.

Displaying a long long value in C++ is straightforward:

long long num = some_value;
std::cout << "num = " << num << "\n";

Or if you prefer printf for some reason, use the "%lld" format for long long, "%llu" for unsigned long long.

(For integers too wide to fit in 64 bits, there are software packages that handle arbitrarily large integers; the most prominent is GNU's GMP. But you don't need it for 15-digit integers.)

Upvotes: 1

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