A small conclusion about GCD and LCM

First, we will introduce two:

GCD with large numbers

Stein + Euclidean

　　

function stein(a,b:int64):int64;begin      if a<b then exit(stein(b,a));      if b=0 then exit(a);      if ((a and 1)=0) and ((b and 1)=0) then exit(stein(a>>1,b>>1)<<1);      if (a and 1)=0 then exit(stein(a>>1,b));      if (b and 1)=0 then exit(stein(a,b>>1));      exit(stein((a+b)>>1,(a-b)>>1));end;

View code

Decimal GCD

Phase Division

　　

function stein(a,b:int64):int64;begin  if a<b then exit(stein(b,a));  if b=0 then exit(a);  if ((a and 1)=0) and ((b and 1)=0) then exit(stein(a>>1,b>>1)<<1);  if (a and 1)=0 then exit(stein(a>>1,b));  if (b and 1)=0 then exit(stein(a,b>>1));  exit(stein((a+b)>>1,(a-b)>>1));end;

View code

 

We often need to calculate the LCM. We have a very elegant conclusion.

A * B/gcd (a, B) = lcm (A, B)

Therefore, we only need to calculate GCD. When A and B are large, it is a good optimization.

Next let's take a look at the question

Description

 

Input

For each test point:
The first row contains an integer T, indicating the number of data groups.
For each group of data that follows, there are two rows.
The first line is an integer n representing the sequence length.
The second line represents the N integers AI separated by spaces, representing the weights calculated for each material.

 

Output

For group I data, you need to specify the number of output groups as "case I:", where I indicates the current number of data groups.
Then, the α and β parameters to be calculated must be output and separated by spaces.
If the required substring does not exist, the corresponding α or β parameter is set to-1.

 

Sample input 327 242 2 3 432 2 4 sample output Case 1: 2 2 Case 2: 4 2 case 3:-1-1 prompt

 

 

This probably means that you need to calculate the two largest substrings respectively, so that gcd (Al, A2 ,...., AR) = 1 lcm (Al ,..... AR) = Al *.... * AR;

GCD is easy to do, and gcd (A [I], a [I + 1]) is continuously read. If gcd (A [I], a [I + 1]) exists, = 1 indicates the interlace, that is, Ans: = N; otherwise, no solution is available.

2. DP practices

F [I] = max (F [I-1] + 1, I-k + 1); k is the number of the last number not equal to the number of ai counterparts.

The answer is max (F [1]..., F [n-1], F [N]);

Complexity O (N)

Solution 2: Maintain the queue
1. maintain such a queue to ensure the mutual quality of the queues.

2. Enable elements to join the queue from left to right. If no mutual quality exists in the queue, enable the first line of the queue until the mutual quality is met. In this process, record the number of elements.

　　　

 

 

A small conclusion about GCD and LCM