Manthan, Codefest 19 (open for everyone, rated, Div. 1 + Div. 2) F. Bits And Pieces sosdp

F. Bits And Pieces

題面

You are given an array 𝑎 of 𝑛 integers.

You need to find the maximum value of 𝑎𝑖|(𝑎𝑗&𝑎𝑘) over all triplets (𝑖,𝑗,𝑘) such that 𝑖<𝑗<𝑘.

Here & denotes the bitwise AND operation, and | denotes the bitwise OR operation.

Input

The first line of input contains the integer 𝑛 (3≤𝑛≤106), the size of the array 𝑎.

Next line contains 𝑛 space separated integers 𝑎1, 𝑎2, ..., 𝑎𝑛 (0≤𝑎𝑖≤2⋅106), representing the elements of the array 𝑎.

Output

Output a single integer, the maximum value of the expression given in the statement.

Examples

input

3
2 4 6

output

6

input

4
2 8 4 7

output

12

Note

In the first example, the only possible triplet is (1,2,3). Hence, the answer is 2|(4&6)=6.

In the second example, there are 4 possible triplets:

(1,2,3), value of which is 2|(8&4)=2.
(1,2,4), value of which is 2|(8&7)=2.
(1,3,4), value of which is 2|(4&7)=6.
(2,3,4), value of which is 8|(4&7)=12.
The maximum value hence is 12.

題意

給你n個數，而後讓你找到三個不一樣的ijk，使得a[i]|(a[j]&a[k])最大

題解

比較顯然的作法是枚舉a[i]，而後從高位到地位看最大的(a[j]&a[k])是多少。

(a[j]&a[k])這個東西咱們單獨維護，枚舉a[j]的每一位，舉個例子好比a[j]是10001，那麼咱們讓cnt[10000],cnt[10001],cnt[00001]都加1；好比00101，咱們讓cnt[00100]和cnt[00001],cnt[00101]都加上1。

而後咱們從高位到地位for循環貪心找到cnt[x]>2的最大的便可。

這個枚舉能夠用sosdp來作；也能夠比較牛逼的操做每一位來枚舉，看法法1。

代碼

#include<bits/stdc++.h>
using namespace std;
const int N=4000005;
int n,cnt[N],ans,a[N];
void insert(int x,int y){
    if (cnt[x|y]==2)return;
    if (x==0){
        cnt[y]++;
        return;
    }
    insert(x&x-1,y|x&-x);
    insert(x&x-1,y);
    // x&x-1是取消最小的一位，x&-x是取最小的一位
}
int main() {
    scanf("%d",&n);
    for (int i=1;i<=n;i++)scanf("%d",&a[i]);
    for (int i=n;i>=1;i--){
        if (i+2<=n){
            int now=0;
            for (int j=20;j>=0;j--)
                if (!((1<<j)&a[i])&&cnt[now|1<<j]==2)now|=1<<j;
            ans=max(ans,now|a[i]);
        }
        insert(a[i],0);
    }
    printf("%d\n",ans);
    return 0;
}

/* sos dp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1<<21;
int dp[maxn][21],a[maxn],n;
void sosdp(int num,int k){
    if(k>20)return;
    if(dp[num][k]>1)return;
    dp[num][k]++;
    sosdp(num,k+1);
    if(num>>k&1){
        sosdp(num^(1<<k),k);
    }
}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
    }
    int ans = 0;
    for(int i=n;i>=1;i--){
        int res = 0, t = 0;
        for(int j=20;j>=0;j--){
            if(a[i]>>j&1){
                res|=1<<j;
            }else if(dp[t|(1<<j)][20]>1){
                res|=1<<j;
                t|=1<<j;
            }
        }
        sosdp(a[i],0);
        if(i<=n-2){
            ans=max(ans,res);
        }
    }
    cout<<ans<<endl;
}
*/