題目大意:

　　現有$n(n \leq 10^{6})$個int範圍內正整數,現欲取其中二數進行與操作,請求最大結果.

solution:

　　欲解此題,必先熟悉位運算.位運算,玄學者也.現有一神奇解法,觀碼以解之.

code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
#include<ctime>
#include
 
<map>
#define R register
#define next exnttttttttt
#define debug puts("mlg")
using namespace std;
typedef int ll;
typedef long double ld;
typedef unsigned long long ull;
inline ll read();
inline void write(ll x);
inline void writesp(ll x);
inline void writeln(ll x);
ll n;
ll num[1100000];
ll ans;
int
 
 main(){
    freopen("sword.in","r",stdin);
    freopen("sword.out","w",stdout);
    n=read();    
    for(R ll i=1;i<=n;i++)num[i]=read();
    for(R ll i=30;i>=0;i--){
        ll t=0;
        for(R ll j=1;j<=n;j++) if(((ans&num[j])==ans)&&(num[j]&(1<<i))) ++t;
        if
 
(t>=2) ans+=(1<<i);
    }
    writeln(ans);
}
inline ll read(){ll x=0,t=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') t=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*t;}
inline void write(ll x){if(x<0){putchar('-');x=-x;}if(x<=9){putchar(x+'0');return;}write(x/10);putchar(x%10+'0');}
inline void writesp(ll x){write(x);putchar(' ');}
inline void writeln(ll x){write(x);putchar('\n');}