cf 156.div2 C Almost Arithmetical Progression

     貌似是两周前的比赛，一直忙考试没写。

     C题就是找个最长的交替序列，直接模拟可以过，比赛的时候以为会超时就没写，刚写了下居然过了

     也可以用dp，dp[i][j]=dp[j][last]+1，last为最后一个和i相等的数下标，就是i和last是两个相同的数，中间加一个j为任意数

模拟代码：

/*
author:jxy
lang:C/C++
university:China,Xidian University
**If you need to reprint,please indicate the source**
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <map>
#include <algorithm>
#define INF 1E9
using namespace std;
map<int,vector<int> > pos;
vector<int> all;
int main()
{
    int n,i,j,t,ans;
    while(~scanf("%d",&n))
    {
        ans=0;
        pos.clear();all.clear();
        for(i=0;i<n;i++)
        {
            scanf("%d",&t);
            if(pos[t].size()==0)all.push_back(t);
            pos[t].push_back(i);
        }
        int a,b,now,A,B,t;
        bool flag=0;
        sort(all.begin(),all.end());
        for(i=0;i<all.size();i++)
        {
            A=all[i];
            ans=max(ans,(int)pos[A].size());
            for(j=i+1;j<all.size();j++)
            {
                B=all[j];
                a=b=0;now=2;
                while(a<pos[A].size()||b<pos[B].size())
                {
                    t=pos[A][a]-pos[B][b];
                    flag=1;
                    while(t*(pos[A][a]-pos[B][b])>0)
                    {
                        if(t<0){a++;if(a==pos[A].size()){flag=0;break;}}
                        else {b++;if(b==pos[B].size()){flag=0;break;}}
                    }
                    if(flag)now++;
                    else break;
                }
                ans=max(now,ans);
            }
        }
        printf("%d\n",ans);
    }
}

dp代码：

/*
author:jxy
lang:C/C++
university:China,Xidian University
**If you need to reprint,please indicate the source**
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#define INF 1E9
using namespace std;
int a[4001];
int dp[4001][4001];
int main()
{
    int i,j,last,n,ans;
    while(~scanf("%d",&n))
    {
        ans=0;
        memset(dp,0,sizeof(dp));
        for(i=1;i<=n;i++) scanf("%d",&a[i]);
        for(i=1;i<=n;++i)
         for(j=last=0;j<i;++j)
         {
             dp[i][j]=dp[j][last]+1;
             if(a[i]==a[j])last=j;
             ans=max(dp[i][j],ans);
         }
        printf("%d\n",ans);
    }
}