HDU-4737 A Bit Fun 维护

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4737

　　题意：给一个数列a₀, a₁ ... , a_n-1，令 f(i, j) = a_i|a_i+1|a_i+2| ... | a_j，求数列中有多少对f(i,j)满足f(i,j)<m。

　　转化为二进制数，依次枚举j，那么只要找到第一个满足的 i 就可以了，我们用个数组w[k]标记每位二进制中的1在j左边第一次出现的位置，然后依次根据w[k]数组中的位置从大到小加数，直到大于m为止，就是此时j对应的最小的i。复杂度O(32*n)。

 //STATUS:C++_AC_1312MS_848KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e60;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 //End

 #define get(a,i) ((a)&(1<<(i)))

 struct Node{
     int l,j;
     bool operator < (const Node& a)const{
         return l>a.l;
     }
 }low[N];

 int l[N];
 int T,n,m;

 int main(){
  //   freopen("in.txt","r",stdin);
     int i,j,b,d,num[N],ok,w,t,ca=;
     LL ans;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&m);
         mem(l,);
         ans=;
         for(i=;i<=n;i++){
             scanf("%d",&b);
             for(j=;j<=;j++){
                 if(get(b,j))l[j]=i;
                 low[j].l=l[j];
                 low[j].j=j;
             }
             if(b>=m)continue;
             sort(low,low+);
             int sum=;
             for(j=;j<=;j++){
                 sum|=low[j].l?<<low[j].j:;
                 if(sum>=m)break;
             }
             ans+=j<=?i-low[j].l:i;
         }

         printf("Case #%d: %I64d\n",ca++,ans);
     }
     return ;
 }