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)。
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 }
