Hdu-5328 Problem Killer

[align=left]Problem Description[/align]
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  

[align=left]Input[/align]
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
 

[align=left]Output[/align]
For each test case, output one line with a single integer, representing the answer.
 

[align=left]Sample Input[/align]

2
5
1 2 3 4 6
10
1 1 1 1 1 1 2 3 4 5

 

[align=left]Sample Output[/align]

4
6

 

[align=left]Author[/align]
XJZX
 

[align=left]Source[/align]
2015 Multi-University Training Contest 4

 

[align=left]Recommend[/align]
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分析：我的写法挺容易写错的，循环中间break的情况要特殊注意边界。

 

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int t,n;
double a[1000006];
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(int i = 1;i <= n;i++) scanf("%lf",&a[i]);
int s = 1,ans = 1;
while(s < n)
{
double d = a[s+1] - a[s];
int i = s+1;
for(;i <= n;i++)
if(a[i] - a[i-1] != d) break;
if(i > n) i--;
if(a[i] - a[i-1] != d) i--;
ans = max(ans,i-s+1);
s = i;
}
s = 1;
while(s < n)
{
double q = a[s+1]/a[s];
int i = s+1;
for(;i <= n;i++)
if(a[i]/a[i-1] != q) break;
if(i > n) i--;
if(a[i]/a[i-1] != q) i--;
ans = max(ans,i-s+1);
s = i;
}
printf("%d\n",ans);
}
}