Project Euler: Problem 57 – Square root convergents

Problem 57:

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?

Idea:

This problem was deceptively simple: just continue the expansion 1000 times, counting how many times there are more digits in the numerator than the denominator. After many attempts, I couldn’t figure out an iterative way, but then I realized this is a powerful example of recursion, and the “deceptively” part vanished and became simple.
 
I wrote a base “root2ConvergenceFraction” that would give me the numerator and denominator (BigInteger form, just to be safe) after n number of expansions (the recursion). Did that from 1 to 1000 and counted the number of instances when the numberator had more digits than the denominator.

int answer = 0;

BigInteger[] tmp;
for(int i = 1; i < = 1000; i++) {
   tmp = EulerUtils.root2ConvergenceFraction(i);
   if(tmp[0].toString().length() > tmp[1].toString().length()) {
      answer++;
   }
}

Project Euler: Problem 31 – Coin sums

Problem 31:

In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:

1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).

 
It is possible to make £2 in the following way:

1x£1 + 1x50p + 2x20p + 1x5p + 1x2p + 3x1p

 
How many different ways can £2 be made using any number of coins?

Idea:

I had no idea how I was going to go about this, so I started researching a similar problem I did a while back in college (how many ways can you make change for a dollar?). Stumbled upon Integer Partitions.

int answer = -1;

int[] denomination = {1,2,5,10,20,50,100,200};

answer = integerPartition(200,denomination.length-1,denomination);

System.out.println("Answer: "+answer);
//==================================
private static int integerPartition(int target, int index, int[] denoms) {
   if(target == 0) {
      return 1;
   }
   if(target < 0) {
      return 0;
   }
   if(index < 0 && target > 0) {
      return 0;
   }
   return integerPartition(target,index-1,denoms) +
          integerPartition(target-denoms[index], index,denoms);
}

In my research, I found this is a form of Integer Partition and came across this link.