Problem 57 : Square root convergents

It is possible to show that the square root of two can be expressed as an infinite continued fraction.

√ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + … ))) = 1.414213…

By expanding this for the first four iterations, we get:

1 + 1/2 = 3/2 = 1.5
1 + 1/(2 + 1/2) = 7/5 = 1.4
1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666…
1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379…

The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?

C++程式碼

#include <iostream>
#include <vector>
#include <iterator>
#include <cassert>
    
using namespace std;

//#define UNIT_TEST

class PE0057
{
private:
    vector<int> getNewNumerator(vector<int>& tmp_den_vec, 
                                int multiple,
 
 vector<int>& tmp_num_vec);
public:
    int getNumberOfFractions(int expansions);
};

vector<int> PE0057::getNewNumerator(vector<int>& tmp_den_vec,
                                    int multiple, vector<int>& tmp_num_vec)
{
    int size_d = tmp_den_vec.size();
    int size_n = tmp_num_vec.size();
    vector<int> new_num_vec;

    if (size_d > size_n)
    {
        for(int i=0; i<size_n; i++)
        {
            new_num_vec.push_back(multiple*tmp_den_vec[i] + tmp_num_vec[i]);
        }
        for(int i=size_n; i<size_d; i++)
        {
            new_num_vec.push_back(multiple*tmp_den_vec[i]);
        }
    }
    else if(size_d < size_n)
    {
        for(int i=0; i<size_d; i++)
        {
            new_num_vec.push_back(multiple*tmp_den_vec[i] + tmp_num_vec[i]);
        }
        for(int i=size_d; i<size_n; i++)
        {
            new_num_vec.push_back(multiple*tmp_den_vec[i]);
        }
    }
    else if (size_d == size_n)
    {
        for(int i=0; i<size_n; i++)
        {
            new_num_vec.push_back(multiple*tmp_den_vec[i] + tmp_num_vec[i]);
        }
    }

    int size_num = new_num_vec.size();
    for(int i=0; i<size_num; i++)
    {
        if (new_num_vec[i] >= 10 )
        {
            if (i==size_num-1)
            {
                new_num_vec.push_back(new_num_vec[i] / 10);
            }
            else
            {
                new_num_vec[i+1] += new_num_vec[i] / 10;
            }
            new_num_vec[i] %= 10;
        }
    }
    return new_num_vec;
}
    
int PE0057::getNumberOfFractions(int expansions)
{
    int numOfFractions = 0;

    vector<int> num_vec; // new numerator digits vector
    vector<int> den_vec; // new denominator digits vector

    vector<int> tmp_num_vec; // temporary numerator digits vector 
    vector<int> tmp_den_vec; // temporary denominator digits vector

    tmp_num_vec.push_back(0);  // initial numerator value 0
    tmp_den_vec.push_back(1);  // initial denominator value 1
 
    for (int i=1; i<=expansions; i++)
    {
        // num_vec/den_vec = 2 + tmp_num_vec/tmp_den_vec 
        //                 = (2*tmp_den_vec + tmp_num_vec)/tmp_den_vec
        // num_vec = 2*tmp_den_vec + tmp_num_vec
        // den_vec = tmp_den_vec 
        num_vec = getNewNumerator(tmp_den_vec, 2, tmp_num_vec);
        den_vec = tmp_den_vec;                             
        
        // change num_vec/den_vec to tmp_den_vec/tmp_num_vec
        tmp_den_vec = num_vec; 
        tmp_num_vec = den_vec;                   

        // num_vec/den_vec = 1 + tmp_num_vec/tmp_den_vec
        //                 = (tmp_den_vec + tmp_num_vec)/tmp_den_vec
        // num_vec = tmp_den_vec + tmp_num_vec
        // den_vec = tmp_den_vec
        num_vec = getNewNumerator(tmp_den_vec, 1, tmp_num_vec); 
        den_vec = tmp_den_vec;

#ifdef UNIT_TEST
        if (i > expansions - 2)
        {
            cout << i << "th expansion, numerator: " ;
            copy(num_vec.rbegin(), num_vec.rend(),ostream_iterator<int>(cout));

            cout << "(" << num_vec.size() << ")" << "    denominator : ";
            copy(den_vec.rbegin(), den_vec.rend(),ostream_iterator<int>(cout));
            
            cout << "(" << den_vec.size() << ")" << endl;
        }
#endif 
        if (num_vec.size() > den_vec.size())
        {
            numOfFractions ++;
        }
    }

    return numOfFractions;
}

int main()
{
    PE0057 pe0057;

    assert (1 == pe0057.getNumberOfFractions(8));

    int numOfFractions = pe0057.getNumberOfFractions(1000);

    cout << "In the first one-thousand expansions, "; 
    cout << numOfFractions << " fractions  " << endl;
    cout << "contain a numerator with more digits than denominator" << endl;

    return 0;
}

Python 程式碼

def getNumberOfFractions(expansions):
    numOfFractions, numerator, denominator = 0, 1, 1

    for i in range(1, expansions+1):
        numerator, denominator = 2*denominator+numerator, denominator+numerator

        #if i > expansions - 2:
        #    print("%dth"%i,"expansion, numerator: ", numerator, end='')
        #    print("(", len(str(numerator)), ")    denominator : ", end='')   
        #    print(denominator, "(", len(str(denominator)), ")")

        if len(str(numerator)) > len(str(denominator)):
            numOfFractions += 1

    return numOfFractions
        
def main():
    assert 1 == getNumberOfFractions(8)
    numOfFractions = getNumberOfFractions(1000)
    print("In the first one-thousand expansions,",numOfFractions,end=' ')
    print("fractions\ncontain a nuerator with more digits than denominator.")

if  __name__ == '__main__':
    main()