D. "Or" Game
time limit per test
memory limit per test
input
output
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
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.
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
.
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
选二进制最高位最大的所有数把他们分别乘以k个x，观察结果
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <map>
#include <set>
#include <queue>
#define MOD 10007
#define maxn 200005
using namespace std;
typedef long long ll;
ll num[maxn], p1[maxn], p2[maxn];
int main(){
int n, k, x;
scanf("%d%d%d", &n, &k, &x);
for(int i = 1; i <= n; i++){
scanf("%I64d", num+i);
p1[i] = p1[i-1] | num[i];
}
for(int i = n; i >= 1; i--)
p2[i] = p2[i+1] | num[i];
ll ans = 0;
for(int i = 1; i <= n; i++){
ll d = num[i];
for(int j = 0; j < k; j++)
d *= x;
ll p = d | p1[i-1] | p2[i+1];
ans = max(ans, p);
}
printf("%I64d\n", ans);
return 0;
}