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Problem Killer

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

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

2
5
1 2 3 4 6
10
1 1 1 1 1 1 2 3 4 5

4
6

暴力

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int a[MAXN],b[MAXN];
double c[MAXN];	
int main()
{
//	freopen("B.in","r",stdin);
	
	int n,T;
	cin>>T;
	while (T--)
	{
		cin>>n;
		For(i,n) scanf("%d",&a[i]);
		int ans=min(2,n);
		For(i,n-1) b[i]=a[i]-a[i+1];
		int p=2;
		Fork(i,2,n-1) if (b[i]==b[i-1]) ans=max(ans,++p);else p=2; 
		
		For(i,n-1) c[i]=(double)a[i]/(double)a[i+1];
		p=2;
		Fork(i,2,n-1) if (fabs(c[i]-c[i-1])<1e-8) ans=max(ans,++p);
		else p=2; 
		
		
		
		cout<<ans<<endl;
		
	}
	
	return 0;
}

HDU 5328(Problem Killer-暴力)

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原文地址：http://blog.csdn.net/nike0good/article/details/47813557