cf 156.div2 C Almost Arithmetical Progression

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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:jxylang: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 1E9using 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:jxylang: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 1E9using 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);    }}