Problem Description
You are a "Problem Killer", you want to solve many problems.
Now you have n problems, the i-th problem's difficulty is represented by an integer ai (1≤ai≤109).
For some strange reason, you must choose some integer l and r (1≤l≤r≤n), and solve the problems between the l-th and the r-th, and these problems' difficulties must form an AP (Arithmetic Progression) or a GP (Geometric Progression).
So how many problems can you solve at most?
Now you have n problems, the i-th problem's difficulty is represented by an integer ai (1≤ai≤109).
For some strange reason, you must choose some integer l and r (1≤l≤r≤n), and solve the problems between the l-th and the r-th, and these problems' difficulties must form an AP (Arithmetic Progression) or a GP (Geometric Progression).
So how many problems can you solve at most?
You can find the definitions of AP and GP by the following links:
https://en.wikipedia.org/wiki/Arithmetic_progression
https://en.wikipedia.org/wiki/Geometric_progression
Input
The first line contains a single integer T, indicating the number of cases.
For each test case, the first line contains a single integer n, the second line contains n integers a1,a2,⋯,an.
For each test case, the first line contains a single integer n, the second line contains n integers a1,a2,⋯,an.
T≤104,∑n≤106
Output
For each test case, output one line with a single integer, representing the answer.
Sample Input
2
5
1 2 3 4 6
10
1 1 1 1 1 1 2 3 4 5
Sample Output
4
6
题意：输入n个数，求其中最长的区间的长度（区间满足等差数列或者等比数列）；
思路：两重循环暴力从每个数开始找，但是这样会超时，可以优化一下，记录从当前数开始的等差数列最远的位置，下次从上次记录的最远的位置开始找，这样复杂度会下降很多；
我发现了一个很有意思的事情，我开始写的程序超时了，后来我删掉了一些东西AC了，但我感觉有bug，于是我想了组数据n=5 5个数4 ，8 ，12 ， 18 ， 27 后面4个数满足等比数列，最长区间应该是4，而我的程序输出是3， 同样这组数据n=6 6个数8 ，16 ，24 ， 36 ， 54 ， 81 后面5个数满足等比数列，输出应该是5，而我的程序输出是4。 我想了一下每次从上次记录的最远的位置开始找，有可能记录的位置的前一位也满足等比数列，所以在计算等比数列长度时得计算一下前一位是否满足；
开始的代码如下：
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <cstring>
using namespace std;
long long a[]; int main()
{
int T;
cin>>T;
while(T--)
{
int n;
scanf("%d",&n);
for(int i=;i<n;i++)
scanf("%lld",&a[i]);
int tmp=;
int t,j;
for(int i=;i<n;i=t)
{
for(j=i+;j<n;j++)
{
if(a[j]!=a[i]+(a[i+]-a[i])*(j-i))
break;
}
t=j-;
tmp=max(tmp,j-i); for(j=i+;j<n;j++)
{
if(a[j-]*a[i+]%a[i]==&&a[j]==a[j-]*a[i+]/a[i])
continue;
else break;
}
tmp=max(tmp,j-i);
//cout<<"t: "<<t<<endl;
//if(j-1>i)
//t=min(t,j-1);
if(t==n-)break;
}
printf("%d\n",tmp);
}
return ;
}
后来修改过的代码：
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <stdlib.h>
#include <cmath>
#include <cstring>
using namespace std;
long long a[]; int main()
{
int T;
cin>>T;
while(T--)
{
int n;
scanf("%d",&n);
for(int i=;i<n;i++)
scanf("%lld",&a[i]);
int tmp=;
int t,j;
for(int i=;i<n;i=t)
{
for(j=i+;j<n;j++)
{
if(a[j]!=a[i]+(a[i+]-a[i])*(j-i))
break;
}
t=j-;
tmp=max(tmp,j-i); for(j=i+;j<n;j++)
{
if(a[j-]*a[i+]%a[i]==&&a[j]==a[j-]*a[i+]/a[i])
continue;
else break;
}
int f=;
if(a[i-]*a[i+]%a[i]==&&a[i]==a[i-]*a[i+]/a[i])
f=;
tmp=max(tmp,j-i+f); if(t==n-)break;
}
printf("%d\n",tmp);
}
return ;
}