Codeforces 578B "Or" Game (前缀和 + 贪心)

Codeforces Round #320 (Div. 1) [Bayan Thanks-Round]

题目链接：B. "Or" Game

You are given \(n\) numbers \(a_1, a_2, ..., a_n\). You can perform at most \(k\) operations. For each operation you can multiply one of the numbers by \(x\). We want to make \(a_1 | a_2 | ... | a_n\) as large as possible, where \(|\) denotes the bitwise OR.

Find the maximum possible value of \(a_1 | a_2 | ... | a_n\) 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 \(a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9)\).

Output

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

Examples

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 \(1 | 1 | 2 = 3\).

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.

Solution

题意

给定 \(n\) 个数 \(a_1 ... a_n\)，可以进行 \(k\) 次操作，每次可以给任意一个数乘上 \(x\)，求 \(a_1 | a_2 | ... | a_n\) 最大为多少。

题解

贪心 前缀和

因为 \(x \ge 2\)，因此一个数乘上 \(x\) 后二进制位数必然增加一位。

由于或运算是 \(0 | 1\) 时才会使答案增加，因此让 \(k\) 个 \(x\) 乘在同一个数上就行。

计算一下前缀和和后缀和，然后暴力枚举每一个数乘以 \(x^k\)，找到最大值即可。

Code

#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
typedef long long ll;

ll a[maxn], s1[maxn], s2[maxn];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n, k, x;
    cin >> n >> k >> x;
    for(int i = 1; i <= n; ++i) {
        cin >> a[i];
        s1[i] = s1[i - 1] | a[i];
    }
    for(int i = n; i >= 1; --i) {
        s2[i] = s2[i + 1] | a[i];
    }
    ll ans = 0;
    ll tmp = x;
    for(int i = 2; i <= k; ++i) {
        tmp *= x;
    }
    for(int i = 1; i <= n; ++i) {
        ans = max(ans, s1[i - 1] | tmp * a[i] | s2[i + 1]);
    }
    cout << ans << endl;
    return 0;
}