Codeforces Round #320 (Div. 2) [Bayan Thanks-Round] D 前缀后缀维护

链接：戳这里

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.

Examples
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.

题意：

给出n个数，k个操作，每个操作是将ai乘以x，问在最多k次操作后 所有n个数的抑或最大 输出答案

思路：

k次操作，且(2<=x<=8) 那么这个数只可能增大而且越来越大

我们肯定是将这个数值(x^k)直接乘到一个数去  这里应该很好理解

接下来时乘到哪个数上去呢？一开始我直接放在最大的那个数上，发现并不能最优

那么我们就直接On枚举，每个数看看放上去使得答案取max

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include <ctime>#include<queue>#include<set>#include<map>#include<stack>#include<iomanip>#include<cmath>#define mst(ss,b) memset((ss),(b),sizeof(ss))#define maxn 0x3f3f3f3f#define MAX 1000100///#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef unsigned long long ull;#define INF (1ll<<60)-1using namespace std;int n,k,x;ll a[200100],pre[200100],last[200100];int main(){    scanf("%d%d%d",&n,&k,&x);    for(int i=1;i<=n;i++) {        scanf("%I64d",&a[i]);        pre[i]=pre[i-1]|a[i];    }    ll num=1;    while(k--) num*=x;    ll ans=0;    for(int i=n;i>=1;i--) last[i]=last[i+1]|a[i];    for(int i=1;i<=n;i++){        ans=max(ans,num*a[i]|last[i+1]|pre[i-1]);    }    printf("%I64d\n",ans);    return 0;}