Codeforces Round #320 (Div. 2)D. "Or" Game（好题）

D. "Or" Game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given n numbers a₁, a₂, ..., a_n. You can perform at most k operations. For each operation you can multiply one of the numbers by x. We want to make as large as possible, where denotes the bitwise OR.

Find the maximum possible value of after performing at most k operations optimally.

Input

The first line contains three integers n, k and x (1 ≤ n ≤ 200 000, 1 ≤ k ≤ 10, 2 ≤ x ≤ 8).

The second line contains n integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁹).

Output

Output the maximum value of a bitwise OR of sequence elements after performing operations.

Sample test(s)

Input

3 1 2
1 1 1

Output

3

Input

4 2 3
1 2 4 8

Output

79

Note

For the first sample, any possible choice of doing one operation will result the same three numbers 1, 1, 2 so the result is .

For the second sample if we multiply 8 by 3 two times we'll get 72. In this case the numbers will become 1, 2, 4, 72 so the OR value will be 79 and is the largest possible result.

  首先，一定是处理转化为二进制之后位数最多的几个数(连续乘以k次x),因为x至少为2，所以位数最多的几个数乘以x之后一定向左移动一位,肯定比其他的数大

其次，为什么不一定是处理最大的数呢,比如

2 1 2

9 12

如果处理12之后答案是25，而处理9之后答案是30，因为12的二进制左移一位之后会和9的二进制有重合，而9的二进制后移一位则不会有这种情况发生 

所以只要枚举处理这n个数中的哪个就行了,如果暴力的话最小是n*n*logK,

这里用了一点特殊的技巧,处理出了前缀的或和和后缀的或和,这样的话可以在n*logK（使用快速幂）的时间复杂度内完成

#include <map>
#include <set>
#include <stack>
#include <queue>
#include <cmath>
#include <ctime>
#include <vector>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
#define inf -0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define mem(a, b) memset(a, b, sizeof(a))
typedef long long ll;
__int64 a[200100];
int pre[200100];
int nex[200100];

int main(){
    int n,k,x;
    while(scanf("%d%d%d",&n,&k,&x)!=EOF){
        __int64 mx=0;
        mem0(pre);
        mem0(nex);
        int id;
        for(int i=1;i<=n;i++){
            scanf("%I64d",&a[i]);
            pre[i]=pre[i-1]|a[i];
            if(a[i]>mx){
                mx=a[i];
                id=i;
            }
        }
        for(int i=n;i>=1;i--)
            nex[i]=nex[i+1]|a[i];
        __int64 ans=0;
        for(int i=1;i<=n;i++){
            for(int j=0;j<k;j++)
                a[i]*=x;
            ans=max(ans,a[i]|pre[i-1]|nex[i+1]);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}