遇到公式题快去推啊，别傻愣 数学推公式 codechef Cracking the Code

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PROBLEM LINK:

Practice

Contest

DIFFICULTY:

Easy-Medium

PREREQUISITES:

Math

PROBLEM:

Given two sequences a and b and
a recurrence to calculate ai and bi,
we need to calculate a certain value c when
a pair of corresponding terms of a and b are
given.

EXPLANATION:

Subtask 1

For subtask 1, we can simply iterate from i to k (as
the case may be) and use the recurrences to calculate ak and bk.

Now, there are two cases:

When i≤k.

In this case, the given recurrence can be used directly. I.e.,

an+1=x(an+bn)−xy(an−bn)

bn+1=x(an−bn)+xy(an+bn)

When k<i.

In this case, the following recurrence can be derived and used.

an−1=an+bn−y(an−bn)2x+2xy2

bn−1=an−bn+y(an+bn)2x+2xy2

The last step is to calculate 2s.
The calculation must be done in floating type to ensure that the answer doesn't overflow. The in-built exponentiation function in most of the programming languages can do this job easily.

Subtask 2

Let us look at the original recurrences given to us, i.e.,

an+1=x(an+bn)−xy(an−bn)

bn+1=x(an−bn)+xy(an+bn)

What do they tell us? Basically, we are given an+1 and bn+1 in
terms of an and bn.
Let us go a step further and try to calculate an+2 and bn+2 in
terms of an and bn.

For doing this, we first note the following terms:

an+1+bn+1=2xan+2xybn

an+1−bn+1=2xbn−2xyan

Now, we use our original recurrence to calculate an+2 and bn+2.

an+2=x(2xan+2xybn)−xy(2xbn−2xyan)=an(2x2+2x2y2)

bn+2=x(2xbn−2xyan)+xy(2xan+2xybn)=bn(2x2+2x2y2)

Since, x=2√ and y=3√,
thus, (2x2+2x2y2)=16.

This leads us to a very important observation:

an+2=16an

bn+2=16bn

It tells us that if i and k are
both even or both odd, then ak+bk=c(ai+bi),
where c is
a constant. When k−i=2, c=24;
when k−i=4, c=28,
and so on. Thus, c=22(k−i).

When i and k are
of different parities, i.e., one is even and the other is odd, we can first calculate ai+1 and bi+1 and
then calculate ak and bk.

The last part is to calculate 2s,
which has already been explained in subtask 1.

COMPLEXITY:

All the operations take constant time except the use of exponentiation function. The in-built exponentiation function is implimented as O(logN) algorithm
in all the programming languages.

Net complexity: O(logN).

SAMPLE SOLUTIONS:

Author

Tester

Editorialist

#include<bits/stdc++.h>

using namespace std;

int main() {
	int i, k, s;
	scanf("%d%d%d", &i, &k, &s);
	
	int ai, bi;
	scanf("%d%d", &ai, &bi);
	
	double a = ai, b = bi;
	if ((i&1) != (k&1)) {
		i++;
		a = sqrt(2.0) * ((ai + bi) - sqrt(3.0) * (ai - bi));
		b = sqrt(2.0) * ((ai - bi) + sqrt(3.0) * (ai + bi));
	}
	double ans = (a + b) * pow(2.0, 2*(k-i) - s);
	printf("%.10f\n", ans);
}