### niklasb's blog

By niklasb, 5 years ago, ,

Can somebody describe their solution to http://codeforces.com/gym/100519/problem/I ? Somehow I can't shake the feeling that it's enough to always multiply with 2, but I cannot substantiate my intuition to turn this into an algorithm.

• +15

By niklasb, 6 years ago, ,

Earlier this day I came across a delightfully interesting question about an algorithm problem: http://stackoverflow.com/questions/23490944/recreate-the-sequence

Alice has written N consecutive, positive integers on a blackboard. E.g. "99, 100, 101, 102". Bob has erased all digits, but one, from each number, so the sequence now reads e.g. "9, 0, 0, 1".

So for every number in the sequence, Bob leaves over exactly one digit, and the position of that digit can vary between the numbers.

You are given the list of remaining digits and you should reconstruct the smallest starting value (in this case, 99) that could have resulted in the given sequence of digits.

OP claims there exists an algorithm. I suggested an algorithm for a starter, but I'm convinced we can do better. Any ideas?

• +24

By niklasb, 6 years ago, ,

I wrote a post about how to tackle certain problems on numbers using dynamic programming techniques over at Stack Overflow. This includes tasks like

• "Find the sum of integers X with digit sum S, where X <= Y" (Y given)
• "Find the number of palindromic integers between L and R"
• "Enumerate all integers between L and R that only have digits 4 and 7"
• "Find the probability that an integer X uniformly chosen from the range [L,R] has at least 10 common digits with a given number S"

that can all be solved with a very similar idea. Just in case somebody's interested.

• +30

By niklasb, 6 years ago, ,

Good morning ;)

I'm currently improving my geometry code collection and searching for programming tasks involving numerical integration to test my code.

So far, I found the following:

I have to add that most of these can be solved explicitly as well :)

Do you know any similar tasks?

I'm using the following code (which I have not written myself!) which is a simple implementation of the adaptive Simpson's rule and which has served me quite well so far:

double simps(double a, double b) {
return (f(a) + 4*f((a+b)/2) + f(b))*(b-a)/6;
}

double integrate(double a, double b){
double m = (a+b)/2;
double l = simps(a,m), r = simps(m,b), tot=simps(a,b);