Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] D 字首字尾維護

連結：戳這裡

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.

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 and is the largest possible result.

題意：

給出n個數，k個操作，每個操作是將ai乘以x，問在最多k次操作後 所有n個數的抑或最大 輸出答案

思路：

k次操作，且(2<=x<=8) 那麼這個數只可能增大而且越來越大

我們肯定是將這個數值(x^k)直接乘到一個數去  這裡應該很好理解

接下來時乘到哪個數上去呢？一開始我直接放在最大的那個數上，發現並不能最優

那麼我們就直接On列舉，每個數看看放上去使得答案取max

程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n,k,x;
ll a[200100],pre[200100],last[200100];
int main(){
    scanf("%d%d%d",&n,&k,&x);
    for(int i=1;i<=n;i++) {
        scanf("%I64d",&a[i]);
        pre[i]=pre[i-1]|a[i];
    }
    ll num=1;
    while(k--) num*=x;
    ll ans=0;
    for(int i=n;i>=1;i--) last[i]=last[i+1]|a[i];
    for(int i=1;i<=n;i++){
        ans=max(ans,num*a[i]|last[i+1]|pre[i-1]);
    }
    printf("%I64d\n",ans);
    return 0;
}