# project euler – Problem 31

November 8, 2011
By

(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
 ```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

R-bloggers.com offers daily e-mail updates about R news and tutorials on topics such as: Data science, Big Data, R jobs, visualization (ggplot2, Boxplots, maps, animation), programming (RStudio, Sweave, LaTeX, SQL, Eclipse, git, hadoop, Web Scraping) statistics (regression, PCA, time series, trading) and more...

If you got this far, why not subscribe for updates from the site? Choose your flavor: e-mail, twitter, RSS, or facebook...

## Recent popular posts

Contact us if you wish to help support R-bloggers, and place your banner here.

# Never miss an update! Subscribe to R-bloggers to receive e-mails with the latest R posts.(You will not see this message again.)

Click here to close (This popup will not appear again)