What is the logic behind pattern i.e.(ans=(n+1)/2) in question ALICESIE on spoj.
Algorithm_given:
1.Create a list of consecutive integers from N to 2 (N, N-1, N-2, ..., 3, 2). All of those N-1numbers are initially unmarked.
2.Initially, let P equal N, and leave this number unmarked.
3.Mark all the proper divisors of P (i.e. P remains unmarked).
4.Find the largest unmarked number from 2 to P – 1, and now let P equal this number.
5.If there were no more unmarked numbers in the list, stop. Otherwise, repeat from step 3.
Find total number of unmarked numbers.
i know its O(sqrt(n)) solution but answer is expected in O(1),it can found by seeing the common pattern i.e.(N+1)/2
But how to prove it Mathematically
link: ALICESIE
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Given an array A[] with n elements, the task is to find S mod (10^9+7), in which S is the smallest perfect square which is divisible by all the elements A[i] (1<=i<=n) of the given array.
So, the problem is very easy if the value of A[i] and n is small. But in this case, I don't know what to do when A[i] can up to 10^7 and n can up to 10^5. So everybody help me pls!
The smallest integer X which is a multiple of all the A_i is called the least common multiple of the A_i. It's also true that every common multiple of the A_i is divisible by X. So S is divisible by X, or equivalently S is a multiple of X.
The LCM can computed fairly efficiently by the algorithms mentioned in the wikipedia article, but remember our final goal is S, a perfect square, not X. Also, the size of X (and S) is likely to be enormous given the constraints in your problem.
Thus I think the correct approach is to use a modified Sieve of Eratosthenes (or just obtain from some online source a list of primes up to 3163) to completely factor all the A_i simultaneously into their prime power factorizations. Since the A_i < 107 you need only include primes <= 103.5. Now, with each A_i factored into its prime power factorization use the prime factorization method to find the LCM, but still retain this in prime power format, in other words don't yet multiply everything together. Next, scan through each of the powers and add 1 to any odd powers. Now you have the prime power factorization of S. Iterate through these prime powers, multiplying each one into the product and taking the product mod (109+7) at each step.
What's the probability that the distinct number appear in a size-k subset of n distinct numbers?
Let A be our target number, S be some the size-k subset of [1,2,3....n].
What's the probability that A is the one of k numbers in S? Many thanks.
PS: I can draw a condition tree diagram, and find the answer maybe the k/n.
But how can I think about it? Thanks again.
The probability is indeed, as you mentioned, k/n. Think about is this way: Let x be an element of [1,2,...,n]. There are binom(n,k) subsets of size k in total, and there are binom(n-1,k-1) subsets of size k that contain x (because x is chosen and we need to choose another k-1 elements). Hence the probability of x to be contained in S is binom(n-1,k-1)/binom(n,k)=k/n.
I am new to permutations and combinations. I am provided with the number n and I can make a number with the help of n+1 digits which are 0,1,...n. I need to find in how many ways I can make the sum n by putting these n+1 digits in n places.
Like i am having number n=2
then
(a) 0 and 2 (b)1 and 1 (c)2 and 0 .
for a number n=3
then
(a)0,0,3 (b)0,3,0 (c)3,0,0 (d)0,1,2 (e)0,2,1 (f)1,0,2 (g)1,2,0 (h)2,0,1 (i)2,1,0 (j)1,1,1
So in total i have 10 ways to generate sum=3 by using the digits 0,1,2,3.
And also consider i can put these n+1 digits in n places only.
The number of compositions of n into k nonnegative summands is (n+k-1) choose n by the stars-and-bars method. You have k=n, so the count is 2n-1 choose n. Your examples were 3C2=3 and 5C3=10.
Yes, the number of total ways for making the sum=n using n+1 digits is equal to (n+n-1)C(n-1) or you can say (n+n-1)C(n).
I was looking at [Stirling numbers of the second kind], which are the total number of ways to split a set of length n into k non-empty subsets, where order does not matter.(http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html), and was wondering how to write a non-naive algorithm to compute
S(n, k {occurences of each element})
Where
S(6, 3, {1, 2, 3} )
would give the total number of ways a set with 6 elements in which 3 are the same element and a different 2 are another element (and 1 is its unique element) could be split into 3 non-empty sets, ignoring permutations.
There is a recursive formula for regular Stirling numbers of the second kind S(n, k), but unlikely to be a comparable function for multisets.
So what's an algorithm that could calculate this number?
Relevant question on Math.SE here, without a real method to calculate this number.
In this question Getting N random numbers that the sum is M, the object was to generate a set of random numbers that sums to a specific number N. After reading this question, I started playing around with the idea of generating sets of numbers that satisfy this condition
sum(A) == sum(B) && sum(B) == sum(A * B)
An example of this would be
A <- c(5, 5, -10, 6, 6, -12)
B <- c(5, -5, 0, 6, -6, 0)
In this case, the three sums equal zero. Obviously, those sets aren't random, but they satisfy the condition. Is there a way to generate 'random' sets of data that satisfy the above condition? (As opposed to using a little algorithm as in the above example.)
(Note: I tagged this as an R question, but the language really doesn't matter to me.)
You'd need to define the first vector in n-dimensional space, and the 2nd one will have N-2 degrees of freedom (i.e. random numbers) since the sum and one angle are already determined.
The 2nd vector would need to be transformed into N-dimensional space; There are infinitely many transforms that could work, so if you don't care about the probability distribution of the resulting vectors, just choose the one that's most intuitive to you.
There's a nice geometrical interpretation to the first constraint: it constrains the 2nd vector to a (hyper-)plane in N-dimensional space; the 2nd constraint doesn't have a simple geometric interpretation.
check out hyperspherical cooridnates.
You can generate one set completely randomly. And generate randomly all numbers in set B except for two numbers. Since you have two equations you should be able to solve for those two numbers.