P3646 [APIO2015]巴厘岛的雕塑

非常好的二进制dp题

参考了这位大佬的思路：https://www.cnblogs.com/zbtrs/p/8495170.html

#include<iostream>
#include<cstdio>                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            
#include<algorithm>
#include<string>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
typedef long long ll;
const ll N=2e3+10,inf=1e18;
ll n,A,B,len,ans,s[N],f[N][N],g[N];
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
    return x*f;
}
inline void solve1(){
    while(len){
        for(ll i=1;i<=n;i++) for(ll j=1;j<=min(i,B);j++) f[i][j]=0;
        f[0][0]=1;
        for(ll i=1;i<=n;i++){
            for(ll j=1;j<=min(i,B);j++){
                for(int k=0;k<i;k++){
                    if(f[k][j-1]){
                        ll t=s[i]-s[k];
                        if((!(t&(1ll<<(len-1))))&&((ans|(t>>len))==ans)) f[i][j]=1;
                    }
                }
            }
        }
        int flag=0;
        for(int i=A;i<=B;i++) if(f[n][i]){flag=1;break;}
        if(flag) ans=ans<<1;
        else ans=ans<<1|1; 
        len--;
    }
}
inline void solve2(){
    while(len){
        for(int i=1;i<=n;i++) g[i]=inf;
        for(int i=1;i<=n;i++){
            for(int j=0;j<i;j++){
                ll t=s[i]-s[j];
                if((!(t&(1ll<<(len-1))))&&((ans|(t>>len))==ans)) g[i]=min(g[i],g[j]+1);
            }
        }
        if(g[n]>B) ans=ans<<1|1;
        else ans=ans<<1;
        len--;
    }
}
int main(){
    n=read();A=read();B=read();
    for(int i=1;i<=n;i++) s[i]=s[i-1]+read();
    ll t=s[n];len=0;
    while(t){len++;t>>=1;}
    if(A!=1) solve1();
    else solve2();
    return printf("%lld
",ans),0; 
}