codeforces2k19's blog

By codeforces2k19, history, 4 weeks ago, In English,

I came across the solution to the problem:

can anyone explain me what is the approach to get this recursive relation(I am a beginner):

ll fun(ll n)



    return INT_MIN;


    return 0;


    return dp[n];


return dp[n];


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4 weeks ago, # |
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Let $$$f(n)$$$ be the maximal number of ribbon pieces of length $$$a$$$, $$$b$$$, and $$$c$$$, so that their sum is $$$n$$$.

Let's assume that you know the solution. For instance the optimal way to cut a ribbon of length $$$n$$$ is $$$abaacb$$$. If you remove the last piece $$$b$$$, you end up with $$$abaac$$$. Since $$$abaacb$$$ was the optimal solution for $$$n$$$, $$$abaac$$$ must be the optimal solution for $$$n-b$$$ (if there would be a better solution for $$$n-b$$$, you can also improve the solution for $$$n$$$). This means that $$$f(n) = 1 + f(n-b)$$$.

Now back to the problem. Of course at the beginning you don't know the optimal solution. But you know that the last piece of the optima solution is $$$a$$$, $$$b$$$ or $$$c$$$. Therefore you can write $$$f(n) = \max(1 + f(n-a), 1 + f(n-b), 1 + f(n-c))$$$. If the optimal solution has a $$$b$$$ piece at the end, $$$1 + f(b)$$$ will be the biggest one.

Now, to make a recursion stop, you need to define stop condition. Clearly $$$f(0) = 0$$$.

And we want that $$$f(n) = -\infty$$$ if you cannot cut the ribbon in such a way. E.g. if $$$n = 7$$$ and $$$a = 5, b = 6, c = 8$$$, then you want that $$$f(n = 7) = -\infty$$$. To accomplish this, you can just define $$$f(n) = -\infty$$$ if $$$n < 0$$$. If it is not possible to cut the ribbon correctly, then you remove elements from the back and reach a negative number (doesn't matter how you remove them).

To make this solution efficient, apply dynamic programming to it. In this code example they used Top-Down-DP.