http://tenka1-2017.contest.atcoder.jp/tasks/tenka1_2017_d
给定N，K和A1...AN,B1...BN,选取若干个Ai使它们的或运算值小于等于K时，使得对应的ΣBi值最大，求最大值。
其实我不大理解为什么要这么弄。
一个数K，如果X小于它，那么K的二进制中第r位是1，X的第r位可以是0或1；但如果K的第r位是0，X的第r位一定是0。
我只能勉强这样想：
可以先选定一个或运算值的上限tmp，如果(Ai|tmp==tmp)，那么根据或运算性质当前这个Ai是肯等可以选的，由于Bi>0，自然能选的越多越好。但是要是一个一个枚举或运算上限显然不现实。所以要按照K的二进制来枚举，把K中位是1的变为0，前面位不变，后面位全变为1。
codeforce上有人这么写：
Let's constder about the pattern of K = 13:
All of buying things are 0 (0 or 1) (0 or 1) (0 or 1) in binary-representation, 0 — 7 in decimal
All of buying things are 1 0 (0 or 1) (0 or 1) in binary-representation, 8 — 11 in decimal
All of buying things are 1 1 0 (0 or 1) in binary-representation, 12 — 13 in decimal
You can choose any pattern to buying, of above patterns.
Let's constder about an another pattern, K = 22:
All of buying things are 0 (0 or 1) (0 or 1) (0 or 1) (0 or 1) in binary-representation, 0 — 15 in decimal
All of buying things are 1 0 0 (0 or 1) (0 or 1) in binary-representation, 16 — 19 in decimal
All of buying things are 1 1 0 0 (0 or 1) in binary-representation, 20 — 21 in decimal
All of buying things are 1 1 0 0 0 in binary-representation, 22 in decimal
You can choose any pattern to buying, of above patterns.
So, you can divide [1, K] into logK parts (maximum). The complexity is N * logK = O(NlogK).
官方题解：
#include<iostream>
#include<cstdio>
using namespace std;
typedef long long ll;
const int maxn=1e5+;
int a[maxn],b[maxn];
int fac[]; int main()
{
int N,K;
scanf("%d%d",&N,&K);
ll res=;
for(int i=;i<N;i++){
scanf("%d%d",&a[i],&b[i]);
if((a[i]|K)==K) res+=b[i];
}
for(int i=;i<=;i++)
fac[i]=<<i;
for(int i=;i<=;i++){
if(K&fac[i]){
ll cnt=;
int tmp=(K^fac[i])|(fac[i]-);//把当前位置为0，前面位不变，后面位全为1
for(int j=;j<N;j++)
if((a[j]|tmp)==tmp) cnt+=b[j];
res=max(res,cnt);
}
}
cout<<res<<endl;
}