Problem Solving report HDU5328 problem Killer

Problem Solving report HDU5328 problem Killer

Description

You is a "problem Killer", you want to solve many problems.
Now we have Problems, the -th problem ' s difficulty is represented by an integer ( ).
For some strange reason, you must choose some integer and ( ), and solve the problems between the -th and the -th, and these problems ' difficulties must form an AP (arithmetic progression) or a GP (geometric progression).
So what many problems can you solve at the most?

You can find the definitions of APS 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 , indicating the number of cases.
For each test case, the first line contains a single integer , the second line contains Integers .

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Output

For each test case, output one line with a single integer, representing the answer.

Sample Input

Sample Output

title: give you a sequence of numbers to find the longest arithmetic progression and the longest geometric series the length of the elder.
analysis: Sweep from left to right, encountered unable to continue the construction of the linear/geometric series number will update the maximum value, re-start the construction of new. Note that restarting the construct does not need to start with the second number of the previous sequence, because it has no meaning, the difference/scale is not the same after the current number, but rather it should be constructed from the last of the previous sequence as a starting point and with a new difference/scale.
on the code:

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm>using namespace std;const int INF = 0x3f3f3f3f;int Main () {int kase;cin >> kase;while (kase--) {int n;scanf ("%d", &n); int AP = in F;double GP = Inf;int Apnum = 2, Gpnum = 2;int Apmax = 0, Gpmax = 0;int Pre, now;scanf ("%d", &pre); if (n = = 1) {printf ("1\n"); continue;} for (int i = 1; i < n; i++) {scanf ("%d", &now), int aptem = now-pre;double Gptem = (double) now/pre;if (AP! = Apte m) {Apmax = max (Apmax, Apnum); Apnum = 2; AP = Aptem;} else{apnum++;} if (GP! = Gptem) {Gpmax = max (Gpmax, gpnum); Gpnum = 2; GP = Gptem;} else{gpnum++;} Pre = Now;} Gpmax = Max (Gpmax, gpnum); Apmax = Max (Apmax, Apnum);p rintf ("%d\n", Max (Gpmax, Apmax));} return 0;}

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Problem Solving report HDU5328 problem Killer