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)。

  1 //STATUS:C++_AC_1312MS_848KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //#pragma comment(linker,"/STACK:102400000,102400000")
 25 //using namespace __gnu_cxx;
 26 //define
 27 #define pii pair<int,int>
 28 #define mem(a,b) memset(a,b,sizeof(a))
 29 #define lson l,mid,rt<<1
 30 #define rson mid+1,r,rt<<1|1
 31 #define PI acos(-1.0)
 32 //typedef
 33 typedef __int64 LL;
 34 typedef unsigned __int64 ULL;
 35 //const
 36 const int N=40;
 37 const int INF=0x3f3f3f3f;
 38 const int MOD=1000000007,STA=8000010;
 39 const LL LNF=1LL<<60;
 40 const double EPS=1e-6;
 41 const double OO=1e60;
 42 const int dx[4]={-1,0,1,0};
 43 const int dy[4]={0,1,0,-1};
 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 45 //Daily Use ...
 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 56 //End
 57 
 58 #define get(a,i) ((a)&(1<<(i)))
 59 
 60 struct Node{
 61     int l,j;
 62     bool operator < (const Node& a)const{
 63         return l>a.l;
 64     }
 65 }low[N];
 66 
 67 int l[N];
 68 int T,n,m;
 69 
 70 int main(){
 71  //   freopen("in.txt","r",stdin);
 72     int i,j,b,d,num[N],ok,w,t,ca=1;
 73     LL ans;
 74     scanf("%d",&T);
 75     while(T--)
 76     {
 77         scanf("%d%d",&n,&m);
 78         mem(l,0);
 79         ans=0;
 80         for(i=1;i<=n;i++){
 81             scanf("%d",&b);
 82             for(j=0;j<=30;j++){
 83                 if(get(b,j))l[j]=i;
 84                 low[j].l=l[j];
 85                 low[j].j=j;
 86             }
 87             if(b>=m)continue;
 88             sort(low,low+31);
 89             int sum=0;
 90             for(j=0;j<=30;j++){
 91                 sum|=low[j].l?1<<low[j].j:0;
 92                 if(sum>=m)break;
 93             }
 94             ans+=j<=30?i-low[j].l:i;
 95         }
 96 
 97         printf("Case #%d: %I64d
",ca++,ans);
 98     }
 99     return 0;
100 }