i have got only 10pts in NSA by sheer brute force! So, I don't think I could be of much help as of now! But, I do presume it to be a greedy/DP question. Will update you if I find a elegant solution!
can anyone help me question is (https://www.codechef.com/JULY18B/problems/EQUILIBR)
in sample test case:
5
5
the ouput is 187500002.
187500002 = P*Q^-1;
can you find the values of P and Q for this test case.
is there anyone who have done the Facebook Hacker Cup 2018 Qualification Round question no.- 2
the Question is-:
Interception
Consider an N-degree polynomial, expressed as follows:
PN * xN + PN-1 * xN-1 + ... + P1 * x1 + P0 * x0
You'd like to find all of the polynomial's x-intercepts — in other words, all distinct real values of x for which the expression evaluates to 0.
Unfortunately, the order of operations has been reversed: Addition (+) now has the highest precedence, followed by multiplication (*), followed by exponentiation (^). In other words, an expression like ab + c * d should be evaluated as a((b+c)*d). For our purposes, exponentiation is right-associative (in other words, abc = a(bc)), and 00 = 1. The unary negation operator still has the highest precedence, so the expression -2-3 * -1 + -2 evaluates to -2(-3 * (-1 + -2)) = -29 = -512.
Input
Input begins with an integer T, the number of polynomials. For each polynomial, there is first a line containing the integer N, the degree of the polynomial. Then, N+1 lines follow. The ith of these lines contains the integer Pi-1.
Output
For the ith polynomial, print a line containing "Case #i: K", where K is the number of distinct real values of x for which the polynomial evaluates to 0. Then print K lines, each containing such a value of x, in increasing order.
Absolute and relative errors of up to 10-6 will be ignored in the x-intercepts you output. However, K must be exactly correct.
Constraints
1 ≤ T ≤ 200
0 ≤ N ≤ 50
-50 ≤ Pi ≤ 50
PN ≠ 0
Explanation of Sample
In the first case, the polynomial is 1 * x1 + 1 * x0. With the order of operations reversed, this is evaluated as (1 * x)(((1 + 1) * x)0), which is equal to 0 only when x = 0.
In the second case, the polynomial does not evaluate to 0 for any real value x.
@zyan1zyan can you help me in first question Tourist (https://www.facebook.com/hackercup/problem/1632703893518337/)
can you give me hint for this problem