【Codeforces 578B】【貪心】"Or" Game

“Or” Game

Time Limit: 2000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u

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.

Samples

Input1

3 1 2
1 1 1

Output1

3

Input2

4 2 3
1 2 4 8

Output2

79

Hint

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.

Source

Codeforces Round #320 (Div. 1) [Bayan Thanks-Round]

一道貪心的題，因爲是or，所以應該讓高位儘量高，所以所有操作應該都在一個數上面，然後可以用前後綴優化orz（我沒想到寫了個nlogn）優化成O（n）orz

下面放代碼:
寫前後綴實際上要短很多orz：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<cstdlib>
#include<cmath>
#include<ctime>
#include<stack>
#define INF 2100000000
#define ll long long
#define clr(x)  memset(x,0,sizeof(x))
#define clrmax(x)  memset(x,127,sizeof(x))

using namespace std;

inline int read()
{
    char c;
    int ret=0;
    while(!(c>='0'&&c<='9'))
        c=getchar();
    while(c>='0'&&c<='9')
    {
        ret=(c-'0')+(ret<<1)+(ret<<3);
        c=getchar();
    }
    return ret;
}

#define M 200005

int n,k,x;
ll a[M],lg2[40],t=1,ans;

int main()
{
    n=read();k=read();x=read();
    for(int i=1;i<=n;i++)
        a[i]=read();
    for(int i=1;i<=k;i++)
        t*=x;
    for(int i=1;i<=n;i++)
        for(int j=a[i];j;j-=j&(-j))
        {
            int temp=j&(-j);
            temp=log((double)temp+0.5)/log(2.0);
            lg2[temp]++;
        }
    for(int i=1;i<=n;i++)
    {
        ll num=a[i]*t;
        for(int j=a[i];j;j-=j&(-j))
        {
            int temp=j&(-j);
            temp=log((double)temp+0.5)/log(2.0);
            lg2[temp]--;
        }
        ll x=0;
        for(int j=0;j<=32;j++)
            if(lg2[j])x+=1<<j;
        x|=num;
        ans=max(ans,x);
        for(int j=a[i];j;j-=j&(-j))
        {
            int temp=j&(-j);
            temp=log((double)temp+0.5)/log(2.0);
            lg2[temp]++;
        }
    }
    cout<<ans;
    return 0;
}

大概就是這個樣子，如果有什麼問題，或錯誤，請在評論區提出，謝謝。