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

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

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;}