HDU 5328 Problem Killer(2015多校联合)

题目链接

戳我

题目大意

一个序列 \(a\), 求这个序列的 连续子序列 中最长的等差序列或者等比序列

样例解释

2

// T

5

//n

1 2 3 4 6

//n个数, 1,2,3,4是最长等差序列,长度为4

10

//n

1 1 1 1 1 1 2 3 4 5

//前6个1满足等差(比)序列,且长度最长是6

解题思路

暴力查询即可

此题坑点: 等比序列 的 公比 可能时 分数.

代码

//Author LJH
//www.cnblogs.com/tenlee
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#define clc(a, b) memset(a, b, sizeof(a))
using namespace std;

const int inf = 0x3f;
const int INF = 0x3f3f3f3f;
const int maxn = 1e6+5;
const double eps = 1e-6;

int n;
double a[maxn];

int main()
{
int T;
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
for(int i = 0; i < n; i++)
{
scanf("%lf", &a[i]);
}
double ap = a[1] - a[0];
double gp = a[1] / a[0];
int len1 = 1, len2 = 1;
int malen1 = 1, malen2 = 1;
for(int i = 1; i < n; i++)
{
if(abs(a[i] - a[i-1] - ap) < eps) len1++;
else
{
ap = a[i] - a[i-1];
len1 = 2;
}
malen1 = (malen1 > len1 ? malen1 : len1);

if(abs(a[i] / a[i-1] - gp) < eps) len2++;
else
{
gp = a[i] / a[i-1];
len2 = 2;
}
malen2 = (malen2 > len2 ? malen2 : len2);
}
printf("%d\n", malen1 > malen2 ? malen1 : malen2);
}
return 0;
}