codeforces 579D(35/600)

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给定n个数，允许进行k次操作，每次将一个数乘以x，目标是使所有数的位运算OR值最大。输入包含n、k和x的值，以及数列。输出最大OR值。第一例中，无论操作哪个数，结果都是3。第二例中，通过两次将8乘以3，可以达到最大OR值79。策略是优先选择位数最多的数进行乘法操作。

You are given n numbers a1, a2, …, an. 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, …, an (0 ≤ ai ≤ 109).

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

Example
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.

这个东西肯定乘在一个上面
位数越多越好

因为前面这一大堆的合等于高一位的-1
所以不用犹豫

然后前缀后缀一处理就可以了

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
typedef long long ll;
ll tu[200001],qz[200002],hz[200002];
int main()
{
    ll daan = 0;
    int n, k, x;
    cin >> n >> k >> x;
    for (int a = 1; a <= n; a++)scanf("%lld", &tu[a]);
    for (int a = 1; a <= n; a++)qz[a] = tu[a] | qz[a - 1];
    for (int a = n; a >= 1; a--)hz[a] = tu[a] | hz[a + 1];
    ll qw = 1;
    for (int a = 1; a <= k; a++)qw *= x;
    for (int a = 1; a <= n; a++)daan = max(daan, qz[a - 1] | hz[a + 1] | tu[a] * qw);
    cout << daan;
}