hdu 5328 Problem Killer 2015多校联合训练赛4 简单题

Problem Killer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1281    Accepted Submission(s): 483

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

求等比或者等差数列中最长的区间的长度。直接贴解题报告了

http://blog.sina.com.cn/s/blog_15139f1a10102vof7.html

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

long long num[1000001];

int main(){
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i = 0;i < n; i++)
            scanf("%I64d",&num[i]);
        int ans = min(n,2);
        int be = 0;
        for(int i = 2;i < n; i++){
            if(num[i] - num[i-1] == num[i-1] - num[i-2]){
            }
            else be = i-1;
            ans = max(ans,i-be+1);
        }
        be = 0;
        for(int i = 2;i < n; i++){
            if(num[i] * num[i-2] == num[i-1] * num[i-1]){
            }
            else be = i - 1;
            ans = max(ans,i-be+1);
        }
        printf("%d\n",ans);
    }
    return 0;
}