BZOJ4300 绝世好题（动态规划）

　　设f[i][j]为前i个数中所选择的最后一个数在第j位上为1时的最长序列长度，转移显然。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
#define N 100010
int n,a[N],f[N][];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4300.in","r",stdin);
    freopen("bzoj4300.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=;i<=n;i++) a[i]=read();
    for (int i=;i<=n;i++)
    {
        int x=;
        for (int j=;~j;j--) f[i][j]=f[i-][j];
        for (int j=;~j;j--) if (a[i]&(<<j)) x=max(x,f[i][j]);
        for (int j=;~j;j--) if (a[i]&(<<j)) f[i][j]=x+;
    }
    int ans=;
    for (int i=;i<=;i++) ans=max(ans,f[n][i]);
    cout<<ans;
    return ;
}