project euler – 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:

    1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p

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

Recursive algorithm is very straightforward.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

coins <- c(1,2,5,10,20,50,100,200)
findChange <- function(money, numCoins) {
    if (numCoins ==1)
        return(1)
    s <- 0
    for (i in 1:numCoins) {
        remain <- money-coins[i]
        if(remain==0)
            s <- s+1
        if(remain>0)
            s <- s+findChange(remain,i)
    }
    return(s)
}
 
print(findChange(200,length(coins)))

> system.time(source("Problem31.R"))
[1] 73682
   user  system elapsed 
   0.89    0.00    0.89 

Related Posts

• November 1, 2011 — Project Euler-Problem 38 (0)
• November 7, 2011 — project euler-Problem 41 (1)
• November 8, 2011 — project euler – Problem 15 (3)
• November 8, 2011 — project euler – Problem 44 (1)
• November 23, 2012 — project euler: problem 61 (0)