Need Hint On This Problem
For removing tle in NSA, you should precalculate the no. of alphabets that are coming after and before a particular alphabet which you will be going to change by some other alphabet from 'a' to 'z' in order to find minimum answer. This approach is quite feasible.
@akshita77 if si<sj in full string is count=0; then just print 0
else if count>0 then follow as i told before
else if count>0 then follow as i told before
@iamrahul
No i am not getting TLE in 2nd subtask of NMN. i am getting wrong answer in 2nd subtask. i think i am missing some corner test cases. if know then tell me.
you can run above code.
No i am not getting TLE in 2nd subtask of NMN. i am getting wrong answer in 2nd subtask. i think i am missing some corner test cases. if know then tell me.
you can run above code.
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is there anyone who have done the Facebook Hacker Cup 2018 Qualification 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.
Example input -
2
1
1
1
4
9
0
-6
2
-2
Example output -
Case #1: 1
0.0
Case #2: 0
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.
Example input -
2
1
1
1
4
9
0
-6
2
-2
Example output -
Case #1: 1
0.0
Case #2: 0
@zyan1zyan
please can you share your approach for first question
Tourist (https://www.facebook.com/hackercup/problem/1632703893518337/)
please can you share your approach for first question
Tourist (https://www.facebook.com/hackercup/problem/1632703893518337/)
if anyone need help in facebook hacker cup question..ask..i have solved all three question.
there is no data structure involved..only logic tha you have to figure out to solve the problems.
there is no data structure involved..only logic tha you have to figure out to solve the problems.
@zyan1zyan
try to solve the given example by your self..regarding how they giving answer.
if you solve all the four examples then you will find a pattern in it.
look carefully at the point from where answer is changing means at which point it gives false.
try to solve the given example by your self..regarding how they giving answer.
if you solve all the four examples then you will find a pattern in it.
look carefully at the point from where answer is changing means at which point it gives false.
@cppcppcpp
try to make polynomial of 2 degree in the same way as 1 degree is given in example.
and then find at which value of x it will become zero.
and read the statements and every point that is given int the question..all have its use.
when u will find the logic then code for answer is of 3 line only.
try to make polynomial of 2 degree in the same way as 1 degree is given in example.
and then find at which value of x it will become zero.
and read the statements and every point that is given int the question..all have its use.
when u will find the logic then code for answer is of 3 line only.
closed account (iGEqDjzh)
@kashish Can you give me some hint on No Minimum No Maximum ques on codechef. I have given a lot of time but the minimum complexity I have reached is still roughly O(n^3)( by counting the occurrence of each number using combinatorics) is there any further optimization I can do ? Pls help
@kashish you explain little bit more about 2nd facebook hackercup qualifications round round problem. If polynomial equation with degree more then 3 . Then how can we find the roots mean values of x ?
@kashish and anyone who done this
in 2nd question of facebook hacker cup
in second sample test case:
4
9
0
-6
2
-2
so polynomial question should be 9*x^4 - 6*x^2+ 2*x- 2
so expresion can be like 9*x^(4-6)*x^(2+2)*x(1-2)
why it is not become zero when x==0 ?
when we put zero in x then it should be zero ?
where am i wrong ?
in 2nd question of facebook hacker cup
in second sample test case:
4
9
0
-6
2
-2
so polynomial question should be 9*x^4 - 6*x^2+ 2*x- 2
so expresion can be like 9*x^(4-6)*x^(2+2)*x(1-2)
why it is not become zero when x==0 ?
when we put zero in x then it should be zero ?
where am i wrong ?
@ipg 2016069
you are not writing the correct equation..you leaves ^0 at the end
correct equation will be
9*x^((4-6)*x)^((2+2)*x)^((1-2)*x)^0
you are not writing the correct equation..you leaves ^0 at the end
correct equation will be
9*x^((4-6)*x)^((2+2)*x)^((1-2)*x)^0
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is there anyone who have solved the 4th question, please tell how to approach for original constraint.