“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.
Sample Input
Input
3 1 2
1 1 1
Output
3
Input
4 2 3
1 2 4 8
Output
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]
由于取或最大值，那么应使最高位尽量高，那么所有操作应该用在同一个数上。
枚举操作第几个数，前缀后缀优化。
不开long long见祖宗。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<string>
#include<iomanip>
#include<ctime>
#include<climits>
#include<cctype>
#include<algorithm>
#define AUTO "%I64d"
using namespace std;
#define smax(x,tmp) x=max((x),(tmp))
#define smin(x,tmp) x=min((x),(tmp))
#define maxx(x1,x2,x3) max(max(x1,x2),x3)
#define minn(x1,x2,x3) min(min(x1,x2),x3)
typedef long long LL;
const int maxn = 200005;
LL a[maxn],pre[maxn],suf[maxn];
LL exp_x[20];
int n,k,x;
int main()
{
#ifndef ONLINE_JUDGE
freopen("game.in","r",stdin);
freopen("game.out","w",stdout);
#endif
scanf("%d%d%d",&n,&k,&x);
exp_x[0] = 1;
for(int i=1;i<=k;i++) exp_x[i] = exp_x[i-1] * x;
for(int i=1;i<=n;i++) scanf(AUTO,a+i);
for(int i=1;i<=n;i++) pre[i] = pre[i-1] | a[i];
for(int i=n;i>=1;i--) suf[i] = suf[i+1] | a[i];
LL ans = 0;
for(int i=1;i<=n;i++) smax(ans,pre[i-1]|suf[i+1]|(a[i]*exp_x[k]));
printf(AUTO,ans);
return 0;
}
