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

本文探讨了如何通过迭代和递归公式计算数值序列的特定值。针对不同情况，提供了两种计算方法：一种适用于给定索引范围内的直接计算，另一种则通过推导额外的递归公式来处理特定条件下的计算需求。最终，文章阐述了如何利用这些方法来计算序列的最终值，并解释了如何计算特定值的二次项。

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https://s3.amazonaws.com/codechef_shared/download/translated/SEPT15/mandarin/CODECRCK.pdf

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);
}