Hdu-5328 Problem Killer

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?

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.

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

 

Author

XJZX

 

Source

2015 Multi-University Training Contest 4

 

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分析：我的写法挺容易写错的，循环中间break的情况要特殊注意边界。

 

#include 
#include 
#include 
using namespace std;
int t,n;
double a[1000006];
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(int i = 1;i <= n;i++) scanf("%lf",&a[i]);
		int s = 1,ans = 1;
		while(s < n)
		{
			double d = a[s+1] - a[s];
			int i = s+1; 
			for(;i <= n;i++)
			 if(a[i] - a[i-1] != d) break;
			if(i > n) i--;
			if(a[i] - a[i-1] != d) i--;
			ans = max(ans,i-s+1);
			s = i; 
		}
		s = 1;
		while(s < n)
		{
			double q = a[s+1]/a[s];
			int i = s+1; 
			for(;i <= n;i++)
			 if(a[i]/a[i-1] != q) break;
			if(i > n) i--;
			if(a[i]/a[i-1] != q) i--;
			ans = max(ans,i-s+1);
			s = i; 
		}
		printf("%d\n",ans);
	}
 }