HDU 5328 Problem Killer

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Problem Killer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 252    Accepted Submission(s): 97

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

 

Recommend

求最长的等差数列 和最长的等比数列

/* ***********************************************
Author        :
Created Time  :2015/7/30 13:57:52
File Name     :2.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 1000000000
#define maxn 1000000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

bool cmp(int a,int b){
    return a>b;
}
int n;
int a[maxn];
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    int t;
	cin>>t;
	while(t--){
		scanf("%d",&n);
		int d=-INF-INF-10;
		double q=0.0;
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
		}
		int Max=1;
		int MMax=1;
		int x=1,y=1;
		for(int i=2;i<=n;i++){
			int d1=a[i]-a[i-1];
			if(d1==d){
				x++;
				if(x>Max){
					Max=x;
				}
			}
			else {
				d=d1;
				x=1;
			}
		}
		for(int i=2;i<=n;i++){
			double q1=a[i]*1.0/(a[i-1]*1.0);
			if(q1==q){
				y++;
				if(y>MMax){
					MMax=y;
				}
			}
			else{
				q=q1;
				y=1;
			}
		}
		if(n==1)printf("%d\n",1);
		else printf("%d\n",max(Max,MMax)+1);
	}
    return 0;
}