Codeforces Round #320 (Div. 2)D. "Or" Game

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概述

                    D. "Or" Game

                time limit per test2 seconds
                memory limit per test256 megabytes
                inputstandard input
                outputstandard 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.

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 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 .

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.

注意：要预处理不然会超时

#include<iostream>
#include<cstdio>
#include<cstring>
#include <algorithm>
#include<queue>
#include<cmath>
using namespace std;
long long a[200002],b[200002],tt[200002];
int main()
{
    int x,k,n,m,i,j;
    while(scanf("%d%d%d",&n,&k,&x)==3)
    {
        for(i=0; i<n; i++)
        {
            scanf("%I64d",&a[i]);
        }
        long long sum=1,maxn=0,temp;
        for(i=0; i<k; i++)
            sum*=x;
        b[0]=a[0];
        for(i=1;i<n;i++)   ///预处理
        {
            b[i]=b[i-1]|a[i];
        }

        tt[n-1]=a[n-1];
        for(i=n-2;i>=0;i--)  ///预处理
        {
            tt[i]=tt[i+1]|a[i];
        }

        for(i=0;i<n; i++)
        {
            temp=sum*a[i];
            temp=temp|b[i-1]|tt[i+1];
            maxn=maxn>temp?maxn:temp;
        }
        printf("%I64dn",maxn);
    }
    return 0;
}