### Square root convergents题号：57 难度： 5 中英对照

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?

### Code


import java.math.BigInteger;

public final class p057 {

public static void main(String[] args) {
long start = System.nanoTime();
long result = run();
long end = System.nanoTime();
System.out.println(result);
System.out.println((end - start) / 1000000 + "ms");
}

public static long run() {
BigInteger n = BigInteger.valueOf(3);
BigInteger d = BigInteger.valueOf(2);
BigInteger TWO = BigInteger.valueOf(2);
int count = 0;
int i=1;
while(i<=1000){
if(n.toString().length()>d.toString().length()) count++;
BigInteger tmp = n;

153
126ms