Codeforces Round #320 (Div. 2) D. "Or" Game 数学-优快云博客

本文介绍了一个编程问题，即通过执行最多k次操作，每次操作可以选择一个数乘以x，来最大化给定序列的或运算值。文章提供了一种解决思路，即将最大位置尽可能设置为1，并通过将数转换为二进制形式来简化问题。

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 $\text{[math]}$ as large as possible, where $\text{[math]}$ denotes the bitwise OR.

Find the maximum possible value of $\text{[math]}$ 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

input

4 2 3
1 2 4 8

output

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 $\text{[math]}$ .

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.

题意：给你n个数，可以在任意位置乘以k次x；

　　　求n个数最大的或值；

思路：首先显然尽量将最大的位置放1；所以尽量乘以大的数；

　　　将全部拆分成二进制，复杂度N*60

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define esp 0.00000000001
const int N=2e5+10,M=1e6+10,inf=1e9;
ll a[N];
ll flag[N];
ll quickmul(ll x,ll y)
{
    ll ans=1;
    while(y)
    {
        if(y&1)
        ans*=x;
        x*=x;
        y>>=1;
    }
    return ans;
}
void update(ll x,ll gg)
{
    int ji=0;
    while(x)
    {
        flag[ji++]+=(gg)*(x%2);
        x>>=1;
    }
}
ll getans()
{
    ll base=1,sum=0;
    for(int i=0;i<=60;i++)
    {
        if(flag[i])
        sum+=base;
        base*=2;
    }
    return sum;
}
int main()
{
    ll x,y,z,i,t;
    scanf("%I64d%I64d%I64d",&x,&y,&z);
    ll mul=quickmul(z,y);
    for(i=0;i<x;i++)
    scanf("%I64d",&a[i]),update(a[i],1);
    ll ans=0;
    for(i=0;i<x;i++)
    {
        update(a[i],-1);
        update(a[i]*mul,1);
        ans=max(ans,getans());
        update(a[i]*mul,-1);
        update(a[i],1);
    }
    printf("%I64d\n",ans);
    return 0;
}

；

转载于:https://www.cnblogs.com/jhz033/p/5672153.html