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"""#
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### 谜题描述
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Gena loves sequences of numbers. Recently, he has discovered a new type of sequences which he called an almost arithmetical progression. A sequence is an almost arithmetical progression, if its elements can be represented as:
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* a1 = p, where p is some integer;
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* ai = ai - 1 + ( - 1)i + 1·q (i > 1), where q is some integer.
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Right now Gena has a piece of paper with sequence b, consisting of n integers. Help Gena, find there the longest subsequence of integers that is an almost arithmetical progression.
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Sequence s1, s2, ..., sk is a subsequence of sequence b1, b2, ..., bn, if there is such increasing sequence of indexes i1, i2, ..., ik (1 ≤ i1 < i2 < ... < ik ≤ n), that bij = sj. In other words, sequence s can be obtained from b by crossing out some elements.
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Input
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The first line contains integer n (1 ≤ n ≤ 4000). The next line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106).
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Output
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Print a single integer — the length of the required longest subsequence.
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Examples
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Input
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2
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3 5
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Output
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2
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Input
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4
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10 20 10 30
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Output
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3
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Note
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In the first test the sequence actually is the suitable subsequence.
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In the second test the following subsequence fits: 10, 20, 10.
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Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
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```python
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\"\"\"
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// Author : snape_here - Susanta Mukherjee
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\"\"\"
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from __future__ import division, print_function
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import os,sys
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from io import BytesIO, IOBase
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if sys.version_info[0] < 3:
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from __builtin__ import xrange as range
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from future_builtins import ascii, filter, hex, map, oct, zip
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def ii(): return int(input())
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def fi(): return float(input())
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def si(): return input()
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def msi(): return map(str,input().split())
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def mi(): return map(int,input().split())
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def li(): return list(mi())
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def lsi(): return list(msi())
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def read():
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sys.stdin = open('input.txt', 'r')
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sys.stdout = open('output.txt', 'w')
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def gcd(x, y):
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while y:
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x, y = y, x % y
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return x
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def lcm(x, y):
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return (x*y)//(gcd(x,y))
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mod=1000000007
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def modInverse(b,m):
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g = gcd(b, m)
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if (g != 1):
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return -1
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else:
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return pow(b, m - 2, m)
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def ceil2(x,y):
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if x%y==0:
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return x//y
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else:
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return x//y+1
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def modu(a,b,m):
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a = a % m
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inv = modInverse(b,m)
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if(inv == -1):
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return -999999999
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else:
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return (inv*a)%m
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from math import log,factorial,cos,tan,sin,radians,floor,sqrt,ceil
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import bisect
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import random
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import string
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from decimal import *
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getcontext().prec = 50
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abc=\"abcdefghijklmnopqrstuvwxyz\"
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pi=3.141592653589793238
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def gcd1(a):
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if len(a) == 1:
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return a[0]
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ans = a[0]
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for i in range(1,len(a)):
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ans = gcd(ans,a[i])
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return ans
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def mykey(x):
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return len(x)
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def main():
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for _ in range(1):
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n=ii()
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a=li()
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d=dict()
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ind = -1
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for i in a:
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if i in d:
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pass
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else:
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ind += 1
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d[i] = ind
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for i in range(n):
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a[i] = d[a[i]]
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#print(a)
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dp = []
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for i in range(n):
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c = [1]*n
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dp.append(c)
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for i in range(n):
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for j in range(i):
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dp[i][a[j]] = max(1+dp[j][a[i]],dp[i][a[j]])
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ans = 0
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for i in range(n):
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for j in range(n):
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ans = max(ans, dp[i][j])
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print(ans)
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# print(\"Case #\",end=\"\")
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# print(_+1,end=\"\")
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# print(\": \",end=\"\")
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# print(ans)
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# region fastio
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BUFSIZE = 8192
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class FastIO(IOBase):
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newlines = 0
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def __init__(self, file):
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self._fd = file.fileno()
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self.buffer = BytesIO()
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self.writable = \"x\" in file.mode or \"r\" not in file.mode
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self.write = self.buffer.write if self.writable else None
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def read(self):
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while True:
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b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
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if not b:
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break
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ptr = self.buffer.tell()
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self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
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self.newlines = 0
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return self.buffer.read()
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def readline(self):
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while self.newlines == 0:
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b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
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self.newlines = b.count(b\"\n\") + (not b)
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ptr = self.buffer.tell()
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self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
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self.newlines -= 1
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return self.buffer.readline()
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def flush(self):
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if self.writable:
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os.write(self._fd, self.buffer.getvalue())
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self.buffer.truncate(0), self.buffer.seek(0)
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class IOWrapper(IOBase):
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def __init__(self, file):
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self.buffer = FastIO(file)
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self.flush = self.buffer.flush
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self.writable = self.buffer.writable
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self.write = lambda s: self.buffer.write(s.encode(\"ascii\"))
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self.read = lambda: self.buffer.read().decode(\"ascii\")
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self.readline = lambda: self.buffer.readline().decode(\"ascii\")
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def print(*args, **kwargs):
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\"\"\"Prints the values to a stream, or to sys.stdout by default.\"\"\"
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sep, file = kwargs.pop(\"sep\", \" \"), kwargs.pop(\"file\", sys.stdout)
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at_start = True
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for x in args:
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if not at_start:
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file.write(sep)
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file.write(str(x))
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at_start = False
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file.write(kwargs.pop(\"end\", \"\n\"))
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if kwargs.pop(\"flush\", False):
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file.flush()
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if sys.version_info[0] < 3:
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sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
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else:
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sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
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input = lambda: sys.stdin.readline().rstrip(\"\r\n\")
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# endregion
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if __name__ == \"__main__\":
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#read()
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main()
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```
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请完成上述谜题的训练场环境类实现，包括所有必要的方法。
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"""
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from bootcamp import Basebootcamp
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import random
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import re
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from bootcamp import Basebootcamp
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def solve_almost_arithmetic_progression(n, a):
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# 优化后的解题算法支持更高效的验证
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if n <= 1:
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return n
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value_map = {}
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idx = 0
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for num in a:
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if num not in value_map:
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value_map[num] = idx
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idx += 1
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compressed = [value_map[num] for num in a]
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dp = [[1] * idx for _ in range(n)]
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max_len = 1
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for i in range(n):
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for j in range(i):
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prev_val = compressed[j]
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current_val = compressed[i]
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dp[i][prev_val] = max(dp[i][prev_val], dp[j][current_val] + 1)
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max_len = max(max_len, dp[i][prev_val])
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return max_len
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class Calmostarithmeticalprogressionbootcamp(Basebootcamp):
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CASE_TYPES = ['random', 'all_same', 'full_aap', 'alternating', 'minimal']
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def __init__(self, min_n=1, max_n=4000, min_val=1, max_val=10**6):
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self.min_n = max(1, min_n) # 确保符合题目约束n≥1
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self.max_n = min(4000, max_n) # 遵守题目最大限制
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self.min_val = min_val
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self.max_val = max_val
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def case_generator(self):
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case_type = random.choice(self.CASE_TYPES)
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# 特殊处理极小案例
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if case_type == 'minimal':
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n = random.choice([1, 2])
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array = [random.randint(self.min_val, self.max_val) for _ in range(n)]
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if n == 2 and random.random() > 0.5:
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array[1] = array[0] # 50%概率生成全同序列
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else:
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n = random.randint(self.min_n, self.max_n)
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if case_type == 'random':
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array = [random.randint(self.min_val, self.max_val) for _ in range(n)]
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elif case_type == 'all_same':
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val = random.randint(self.min_val, self.max_val)
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array = [val] * n
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elif case_type == 'full_aap':
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p = random.randint(self.min_val, self.max_val)
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q = random.randint(1, self.max_val//2) # 确保q≠0
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array = [p]
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for i in range(2, n+1):
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sign = (-1) ** (i + 1)
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array.append(array[-1] + sign * q)
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elif case_type == 'alternating':
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base = random.sample(range(self.min_val, self.max_val+1), 2)
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array = [base[i%2] for i in range(n)]
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expected_length = solve_almost_arithmetic_progression(n, array)
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return {
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'n': n,
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'array': array.copy(),
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'expected_length': expected_length
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}
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@staticmethod
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def prompt_func(question_case):
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n = question_case['n']
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array = question_case['array']
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return f"""Gena的几乎等差数列定义如下：
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1. 首项a₁是任意整数p
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2. 后续项满足aᵢ = aᵢ₋₁ + (-1)^(i+1)*q（q为整数）
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给定长度为{n}的整数序列：[{', '.join(map(str, array))}]
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请找出其中最长的满足条件的子序列长度
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答案请用[answer]答案[/answer]包裹，例如：[answer]5[/answer]"""
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@staticmethod
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def extract_output(output):
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try:
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matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
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if matches:
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value = matches[-1].strip()
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if '.' in value: # 处理可能的浮点格式
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return int(float(value))
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return int(value)
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except (ValueError, TypeError):
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pass
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return None
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@classmethod
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def _verify_correction(cls, solution, identity):
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try:
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return int(solution) == identity['expected_length']
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except:
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return False
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