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 ... , an-1，令 f(i, j) = ai|ai+1|ai+2| ... | aj，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=40;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-6;
const double OO=1e60;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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
;

int l
;
int T,n,m;

int main(){
//   freopen("in.txt","r",stdin);
int i,j,b,d,num
,ok,w,t,ca=1;
LL ans;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
mem(l,0);
ans=0;
for(i=1;i<=n;i++){
scanf("%d",&b);
for(j=0;j<=30;j++){
if(get(b,j))l[j]=i;
low[j].l=l[j];
low[j].j=j;
}
if(b>=m)continue;
sort(low,low+31);
int sum=0;
for(j=0;j<=30;j++){
sum|=low[j].l?1<<low[j].j:0;
if(sum>=m)break;
}
ans+=j<=30?i-low[j].l:i;
}

printf("Case #%d: %I64d\n",ca++,ans);
}
return 0;
}