hitman623's blog

By hitman623, 3 years ago, In English

Shooting Game

Let PA and PB be the probability of Alice and Bob hitting the target respectively. Then the chances of Alice winning the game are:

PA + (1 - PA)(1 - PB)PA + (1 - PA)2(1 - PB)2PA + ...

which equals using GP formula. Similarly, for Bob it would be .

Equating both we get, . Return -1 in case you get PB greater than 1.

SolveForTrisha

Let x be an nth root of unity. Then, xn = 1. So, xn - 1 = 0 has roots 1, α1, α2....αn - 1. We can write this as follows :

Placing x = 1,  - 1 in the above equation and multiplying them, one can get the desired result.
If n is even the expression evaluates to 0 otherwise n.

Polynomial Guess

Consider the polynomial:

f(x) = x(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)

Let g(r) be the value of at x = r and y(r) be the given values of the polynomial for r = 0 to 5.

The desired polynomial can be written as :

Placing x = 6 in above equation, the desired result can be obtained.

Code

FunctionGame

The main part of the solution here is using the equation : φ(xr) = xr - 1φ(x) , . Rest is just solving GP sums and calculating φ values from 1 to m.

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PrimeMaker

Since, the operations allowed are only modifying / deleting a digit and value of n is at max 106, the maximum number that can be formed is of the rage 10n. The idea is to calculate all primes upto 107 and take the minimum number of operations required to convert n to a prime p over all primes p.

Another solution is to observe that we can always get a prime number in the range n - 99 to n + 99 . So, the maximum value of the answer can be 2 only. So, we just need to check whether the answer is 0 or 1. For answer to be 0, n should be prime. For answer to be 1 we have O(10 * d) candidates only where d is the number of digits in decimal representation of n. Each candidate can be checked in O(sqrt(n)).

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CountingTriangles

Let us try to find a series for the number of triangles after n steps. First of all the side length of the triangle after n steps would be m = 2n. Lets iterate over the side length of the triangles that we are going to count currently. Let it be l. Every triangle is either pointing up or pointing down. Lets count them separately and add them in the end. For counting triangles of length l pointing up, we just need to count the number of valid points that can be the topmost point of our triangle. This is equal to . For counting triangles pointing down, we need to count the valid positions for placing the base of length l. This is equal to . The final answer would be :

.

Simplifying and using some series formulas, you can get a formula as : .
If you do not want to solve the series, observe that the final expression would be of order 3 in m. So, instead assume a general 3 degree polynomial, and get its coefficients using base cases already given.

ConfusingTriangle


The idea here is to use the mirror property. Imagine that the sides of the triangles have mirrors placed on them. Then the shortest distance between two centroids such that the line between them intersects all the sides exactly once is your answer. Applying pythagoras theorum, the answer comes to be

MinimizeExpression

The key to the solution is to use Weighted AM — GM inequality here. Let us write the expression as

. . . . a times . . . . b times . . . c times )

Comparing the RHS of the expression with xAyBzC, solve for a, b and c. The final answer hence would be .
The conditions given guarantee that a, b and c after solving would be positive.

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EquationSolver

The answer to the problem is the coefficient of xc in the expansion of :

(1 + x + x4 + x9 + .....)n

where (1 + x + x4 + x9 + .....) represents the choices each variable has.
One solution is to perform binary exponentiation of the polynomial using NTT for polynomial multiplication. The time complexity of such a solution would be O(clog(c)log(n)).
Another solution is to calculate NTT of the polynomial once, exponentiate individual terms to n and then calculate the inverse NTT. Time complexity of the solution would be O(c(log(c) + log(n))).

Code: 1st solution
Code: 2nd solution

Tribonacci

T(n) = Mn where M is the given matrix.







Let and .

Ti + β)

Let R(n) = T(α)n





The main part of the solution is to calculate and .
Consider the following matrix P(A)
\begin{bmatrix} A & I \\ O & I \end{bmatrix} The matrices P2, P3 . . . are as follows :
\begin{bmatrix} A^{2} & A+I \\ O & I \end{bmatrix}

\begin{bmatrix} A^{3} & A^{2}+A+I \\ O & I \end{bmatrix} Observe that the 2nd element of the first row is the GP sum of matrix A. Therefore,

Now, consider .



The first part of which we already know. For second part, Consider
The matrices S(2), S(3) . . . are as follows :
\begin{bmatrix} A^{2} & A+2I \\ O & I \end{bmatrix}

\begin{bmatrix} A^{3} & A^{2}+2A+3I \\ O & I \end{bmatrix} Observe that the second element of the first row is nothing but the desired part. S(n) can also be calculated in the same way the GP sum was calculated as it is nothing but another GP in P. The overall time complexity of the solution is O(k3(log(n) + log(m))).

Also observe that the modulus is 231, so no need to take modulus. We can let the operations overflow. In the end we just add 231 if result comes out to be negative.
Another method of taking modulus is to take bitwise AND of the intermediate calculations with 231 - 1. Both of them are much faster than the modulo operation.

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3 years ago, # |
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Auto comment: topic has been updated by hitman623 (previous revision, new revision, compare).

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3 years ago, # |
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How do you "Comparing the RHS of the expression with ..." in MinimizeExpression?

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    3 years ago, # ^ |
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    The goal here is to choose such a, b, c (a, b, c > 0) such that the RHS of the expression in the inequality comes to be a function of xAyBzC. For example : given xy2 = 4, to minimize x + y you write it as so that you get xy2 on RHS.

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      3 years ago, # ^ |
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      No, I mean, how to relate the right hand side x^(an1+cm3).. with x^ay^bz^c=s? Or to put it the other way, why should we assume that is s?

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        3 years ago, # ^ |
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        Comparing them, you get the following : .

        You are free to choose any k > 0 and solve for a, b, c. The answer is same .

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3 years ago, # |
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how to upsolve ? can you move them to practice

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    3 years ago, # ^ |
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    I have contacted hmehta. I'll let you know as soon as he replies.

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      3 years ago, # ^ |
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      The match is up in the practice rooms.

      You will find it in SRMs Category under practice rooms as shown below in the image! :)

      Setup Topcoder Java Applet: topcodr.co/SRMGuide

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4 weeks ago, # |
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Hi hitman623,

Would you kindly provide more hints about SolveForTrisha?

When I try to place x = 1,  - 1 in the intermediate equation, the LHS becomes a division-by-zero. So I'm lost there.

Is there something more I should know about this problem? Thanks!