When you make a generalization like: A^n.B^m does that mean that the m can never be equal to n or can I have at least one value where they are equal like "AABB" for example?
This generalization allows for all combinations of m and n including m = n.
Related
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 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
i want to know is it possible to validate that deviding two number has remaining zero in result or not?
for example dividing number 4 on number two has zero in remaining.
4/2=0 (this is true)
but 4/3=1 (this is not true)
is there any expression for validation such case?
Better Question :
Is There any validation expression to validate this sentence ?
Remainder is zero
thank you
You can use a Modulo operator. The modulo operation finds the remainder of division of one number by another
y mod x
5 mod 2 =1 (2x2=4, 5-4=1)
9 mod 3 = 0 (3*3=9)
You can think of it, how many times does x fit in y and then take the remainder.
In computing the modulo operator is integrated in most programming languages, along with division, substraction etc. Check modulo and then your language on google (probably its mod).
This is called the modulo function. It essentially gives the remainder of a division of two integer number. So you can test for the modulo funtion returning zero. For example, in Python you would write
if a % b == 0:
# a can be divided by b with zero remainder
Given two integers is there an easy way to find the largest modulus of congruence for them? i.e. a % n == b %n, Or even to enumerate all of them? Obviously, I could try every value less than them, but it seems like there should be an easier way.
I tried doing something with gcds, but then you just get things where a % n == b % n == 0, which isn't as cool as I was hoping for, and I'm pretty sure this isn't necessarily the largest n.
Any ideas?
If a and b are the numbers, then we consider:
a = nx + r
b = ny + r
Where n is the modulus we want to find, and r is the common remainder. So,
a - b = n(x - y)
Maximum n is achieved with x - y = 1. So,
n = a - b
(If I understood the question correctly.)
I'm not asking about the definition but rather why the language creators chose to define modulus with asymmetric behavior in C++. (I think Java too)
Suppose I want to find the least number greater than or equal to n that is divisible by f.
If n is positive, then I do:
if(n % f)
ans = n + f - n % f;
If n is negative:
ans = n - n % f;
Clearly, this definition is not the most expedient when dealing with negative and positive numbers. So why was it defined like this? In what case does it yield expediency?
Because it's using "modulo 2 arithmetic", where each binary digit is treated independently of the other. Look at the example on "division" here
You're mistaken. When n is negative, C++ allows the result of the modulus operator to be either negative or positive as long as the results from % and / are consistent, so for any given a and b, the expression (a/b)*b + a%b will always yield a. C99 requires that the result of a % b will have the same sign as a. Some other languages (e.g., Python) require that the sign of a % b have the same sign as b.
This means the expression you've given for negative n is not actually required to work in C++. When/if n%f yields a positive number (even though n is negative), it will give ans that's less than n.