标签:... Progression sequence Almost int subsequence integer input DP
C. Almost Arithmetical Progression
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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:
- a1 = p, where p is some integer;
- ai = ai - 1 + ( - 1)i + 1·q (i > 1), where q is some integer.
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.
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.
Input
The first line contains integer n (1 ≤ n ≤ 4000). The next line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106).
Output
Print a single integer — the length of the required longest subsequence.
Examples
input
Copy
2 3 5
output
Copy
2
input
Copy
4 10 20 10 30
output
Copy
3
Note
In the first test the sequence actually is the suitable subsequence.
In the second test the following subsequence fits: 10, 20, 10.
【思路】
dp[i][j] i表示当前数的位置,j表示前面的数。
AC代码:
#include <iostream>
#include <cstring>
using namespace std;
int b[4005];
int dp[4005][4005];
int N;
int main()
{
cin>>N;
for(int i=1;i<=N;i++)
{
cin>>b[i];
}
memset(dp,0,sizeof(dp));
int res=0;
for(int i=1;i<=N;i++)
{
int t=0;
for(int j=0;j<i;j++)
{
dp[i][j]=dp[j][t]+1;
if(b[i]==b[j])
t=j;
res=max(res,dp[i][j]);
}
}
cout<<res<<endl;
return 0;
}
标签:...,Progression,sequence,Almost,int,subsequence,integer,input,DP 来源: https://blog.csdn.net/qq_43956340/article/details/99819234
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