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 (1ai109).
For some strange reason, you must choose some integer l and r (1lrn), 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

T104,n106
 

Output
For each test case, output one line with a single integer, representing the answer.
 

Sample Input
251 2 3 4 6101 1 1 1 1 1 2 3 4 5
 

Sample Output
46
 

Author
XJZX
 

Source
2015 Multi-University Training Contest 4
 

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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;
}


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