project euler – Problem 31
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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
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