HDU5328-Problem Killer

Problem Killer

                                                                                 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                                            Total Submission(s): 2795    Accepted Submission(s): 964

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

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

 

Recommend

wange2014

 

题意：给你n个数，求等比或者等差数列中最长的区间的长度

解题思路：处理出公差和公比序列，然后用尺取法求解

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF=0x3f3f3f3f;int n,a[1000009],b[1000009];double c[1000009];int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            if(i>1)            {                b[i]=a[i]-a[i-1];                c[i]=1.0*a[i]/a[i-1];            }        }        int head=1,rear=1,ma=1;        while(rear<n)        {            rear++;            while(b[head+1]!=b[rear]&&rear>head) head++;            ma=max(ma,rear-head+1);        }        head=1,rear=2;        while(rear<n)        {            rear++;            while(c[head+1]!=c[rear]&&rear>head) head++;            ma=max(ma,rear-head+1);        }        printf("%d\n",ma);    }    return 0;}