Re: [Haskell-cafe] Rewrite this imperative in FP way

"+= a[i]" is the same as "+=1", isn't it? (i accidentally didn't reply to the list on my first try. sorry.) Am 05.02.2012 16:36, schrieb Haisheng Wu:

Sorry there is a mistake in the problem description. Here it is in Python:

a = [1,1,1,1] b = [0,1,2,3] c = [0,2] d = [0,0,0,0] for i in b: for j in c: if (i+j)<3: d[i+j] += a[i]

-Haisheng

On Sun, Feb 5, 2012 at 2:28 PM, Haisheng Wu wrote:

...

a = [1,1,1,1] b = [0,1,2,3] d = [0,0,0,0]

for i in b: for j in c: if (i+j)<3: d[i+j] += a[i]

My just work implementation in Haskell http://hpaste.org/57452

Another people implementation in Haskell with Monad and it turns out complex and very imperatively. http://hpaste.org/57358

Do you have any cool solution in FP way?

Thanks. -Simon

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