project euler – Problem 31

November 8, 2011
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(This article was first published on YGC » R, and kindly contributed to R-bloggers)

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.

?View Code RSPLUS
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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 

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