Project Euler task 57
It is possible to show that the square root of two can be expressed as an infinite continued fraction.
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?
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#include < iostream >
#include < string >
#include "BigIntegerLibrary.hh"
using namespace std;
int main() {
BigInteger iNom = 1;
BigInteger iDen = 2;
BigInteger iTemp = 0;
int iNumber = 0;
for (int i = 0; i < 1000; i++) {
if (bigIntegerToString(iDen + iNom).length() > bigIntegerToString(iDen).length())
iNumber++;
iTemp = iDen;
iDen = iDen * 2 + iNom;
iNom = iTemp;
}
cout << iNumber << endl;
return 0;
}
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4 s 20 ms
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