"Or" Game

"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, ..., 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.ide

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

Input

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

The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 109).code

Output

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

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 .blog

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.three

 首先,這是一道貪心題;瞭解爲何貪心:ip

    二進制位數越多,數字天然越大,x最小爲2,每乘一次x,2進制位數至少加1;因此天然要將指定的一個數乘以(x^k)爲最優;ci

其次,爲何不是直接最大值乘以(x^k):element

    這是或運算特性,渣渣說不上來 = =!舉例 反證 :

                              2 1 2                               12 9
                              24|9 = 25; 12|18 = 30;

 

而後,學會預處理,能夠減小很大的複雜度:

    正着累加一邊pre[i] = pre[i-1] | a[i];

    倒着累加一邊pre2[i] = pre[i+1] | a[i];

最後,依次比較pre[i-1]|(a[i]*x^k)|pre2[i+1],暴力便可。

複雜度爲O(3n);

#include<cstdio>
#include<algorithm>
#define ll __int64
using namespace std;
ll a[200005];
ll pre[200005];
ll pre2[200005];
int main(){
    int n,k,x;
    scanf("%d%d%d",&n,&k,&x);
    int maxn = 0;
    int j = 0;
    for(int i = 1; i <= n; i++){
        scanf("%I64d",&a[i]);

    }
    int op = 1;
    while(k--)
        op *= x;
    ll ans = 0;
    for(int i = 1;i <= n; i++){
        pre[i] = pre[i-1]|a[i];
    }
    for(int i = n; i >= 1; i--){
        pre2[i] = pre2[i+1]|a[i];
    }
    for(int i = 1; i <= n; i++)
        ans = max(ans,pre[i-1]|(a[i]*op)|pre2[i+1]);
    printf("%I64d\n",ans);
    return 0;
}
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