多校4 1002 Problem Killer

Problem Killer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2808    Accepted Submission(s): 605

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

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 1000000 + 5;

int T, n;

int a[N], b[N];

double q[N];

int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        a[0] = 0;
        scanf("%d", &a[1]);
        for(int i = 2; i <= n; i ++){
            scanf("%d", a + i);
            b[i] = a[i] - a[i-1];
        }
        int cnt = 2;
        int ans = 0;
        /*for(int i = 1; i <= n; i ++)
            printf("%d
", b[i]);
        /*for(int i = 1; i <= n; i ++)
            printf("%.2f
", q[i]);*/
        for(int i = 3; i <= n; i ++){
            if(b[i] == b[i-1]){
                cnt += 1;
                continue;
            }
            else{
                ans = max(cnt, ans);
                cnt = 1;
                if(a[i-1] + a[i+1] == 2 * a[i]) cnt += 1;
            }
        }
        ans = max(ans, cnt);
        //printf("ans = %d
", ans);
        cnt = 1;
        for(int i = 2; i < n; i ++){
            if(1ll * a[i] * a[i] == 1ll * a[i-1] * a[i+1]){
                cnt ++;
            }
            else{
                ans = max(ans, cnt + 1);
                cnt = 1;
                if(1ll * a[i] * a[i] == 1ll * a[i-1] * a[i + 1]) cnt += 1;
            }
        }
        if(1ll * a[n-1] * a[n-1] == 1ll * a[n-2] *a[n]) cnt += 1;
        ans = max(ans, cnt);
        if(n == 0) ans = 0;
        if(n == 1) ans = 1;
        if(n == 2) ans = 2;
        printf("%d
", ans);
    }
    return 0;
}