Codeforces Round #320 (Div. 1) [Bayan Thanks-Round] B. "Or" Game 线段树贪心

B. "Or" Game

Time Limit: 1 Sec  

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/578/problem/B

Description

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.

Sample Input

3 1 2
1 1 1

Sample Output

3

HINT

 

题意

给你n个数，你可以操作k次，使得其中的某一个数乘以x

要求最后得到的所有数的或值最大

题解：

首先dp是错的，因为a>b不能保证a|c>b|c

这儿有一个贪心的操作，如果又一次乘法给了a1，那么剩下的都得给a1，这样才能得到最优值

因为x>=2可以保证最高位的1在不断增加

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 2000000 + 500
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
//**************************************************************************************
long long n , k , x;

struct tree
{
    int L , R  ;
    ll sum;
};

tree a[maxn * 4];
ll d[maxn];
void build(int x,int l,int r)
{
    a[x].L = l,a[x].R = r;
    if(l == r)
    {
        a[x].sum = d[l];
        return;
    }
    else
    {
        int mid = (l+r)>>1;
        build(x<<1,l,mid);
        build(x<<1|1,mid+1,r);
        a[x].sum = a[x<<1].sum | a[x<<1|1].sum;
    }
}

ll query(int o,int QL,int QR)
{
    int L = a[o].L , R = a[o].R;
    if (QL <= L && R <= QR) return a[o].sum;
    else
    {
        int mid = (L+R)>>1;
        ll res = 0;
        if (QL <= mid) res |= query(2*o,QL,QR);
        if (QR > mid) res |= query(2*o+1,QL,QR);
        return res;
    }
}
int main()
{
    cin>>n>>k>>x;

    for(int i = 1 ; i <= n ; ++ i)
    {
        scanf("%I64d",&d[i]);
    }
    build( 1 , 0 , n+1);
    ll ans = 0;
    ll temp = 1;
    for(int i=1;i<=k;i++)
        temp *= x;
    for(int i = 1 ; i <= n ; ++ i)
    {
        ans = max( ans , (d[i]*temp) | query(1,0,i-1) | query(1,i+1,n+1));
    }
    cout<<ans<<endl;
}