解题报告 之 HDU5328 Problem Killer

解题报告 之 HDU5328 Problem Killer

Description

You are a "Problem Killer", you want to solve many problems. 
Now you 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 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 , indicating the number of cases. 
For each test case, the first line contains a single integer , the second line contains  integers . 

 

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 

 

题目大意：给你一个数字序列，求其中 最长等差数列 和 最长等比数列 两者中更长者的长度。

分析：从左到右扫一遍，遇到无法继续接着构造等差/等比数列的数字就更新一下最大值，重新开始构造新的。注意重新开始构造不需要从上一个数列的第二个数字开始，因为没有意义，到了当前数字之后 差值/比例还是不一样；而是应该从上一个数列的最后一个作为起点，用新的差值/比例构造。

上代码：

#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 = INF;
		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 != APtem)
			{
				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 );

		printf( "%d\n", max( GPmax, APmax ) );
	}
	return 0;
}