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\n",ca++,ans);

 98     }

 99     return 0;

100 }