Round 1 A-"Or" Game CodeForces

题目链接：
https://vjudge.net/contest/168327#problem/A

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.

Example
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.

大意：
给出n个数字 和数字 x，k 。最多进行 k 次操作，每次操作将一个数字乘上 x ，每个数字OR，求值最大 。

思路：
对于 OR（或）操作和 ai ， ai 的值越大值越大，因而考虑枚举每个数字，每个数字乘上 x k次即可。

#include <bits/stdc++.h>using namespace std;typedef long long ll;#define mem(s,t) memset(s,t,sizeof(s))#define D(v) cout<<#v<<" "<<v<<endl#define inf 0x3f3f3f3f//#define LOCALconst ll MAXN =  2e5+10;ll L[MAXN],R[MAXN],a[MAXN];ll Max(ll a , ll b ){return a>b? a: b;}//不用快速幂直接乘起来就可以过了ll quick_pow(ll a,ll b){    ll ans=1;    ll temp=a;    while(b){        if(b&1)            ans*=temp;        temp*=temp;        b>>=1;    }    return ans;}int main() {#ifdef LOCAL    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);#endif    mem(L,0);    mem(R,0);    mem(a,0);    ll n,k,x;    while(~scanf("%lld%lld%lld",&n,&k,&x)){        ll M =quick_pow(x,k);        for(ll i=1;i<=n;i++){            scanf("%lld",&a[i]);        }        for(ll i=1;i<=n;i++){            L[i]=L[i-1]|a[i];        }        for(ll i=n;i>=1;i--){            R[i]= R[i+1]|a[i];        }        ll ans=-1,temp=0;        //D(quick_pow(x,k));        for(ll i=1;i<=n;i++){            temp=L[i-1]|(M*a[i])|R[i+1];            ans=Max(ans,temp);        }        printf("%lld\n",ans);    }    return 0;}