BZOJ4976 宝石镶嵌（动态规划）

　　显然被留下的宝石应该贡献至少一位，否则就可以扔掉。所以如果n-k>=logw，直接输出所有数的or。现在n变得和k同阶了。于是设f[i][j]为前i个数or为j时至少选几个数，转移显然。当然可以只开一维。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
#define M 120
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
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;
}
int n,m,a[N],f[<<],ans;
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4976.in","r",stdin);
    freopen("bzoj4976.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=n-read();
    for (int i=;i<=n;i++) ans|=a[i]=read();
    if (m>=) {cout<<ans;return ;}
    memset(f,,sizeof(f));
    f[]=;
    for (int i=;i<=n;i++)
        for (int j=;j<(<<);j++)
        f[j|a[i]]=min(f[j|a[i]],f[j]+);
    for (int i=(<<)-;~i;i--)
    if (f[i]<=m) {cout<<i;break;}
    return ;
}