开始写的时候是用的O(n*(logn)^2)的复杂度,枚举起点二分最远能到多远然后累加长度就是答案,但是这样TLE了,最后直接在树上二分完事。
这里顺序是倒着来的,因为这个写法是查最右点,是为了妥协不能区间减法而固定最左点的,所以去右子树的话一定要带上左子树的状态去递归。回来之后还要再max一下,因为有左子树保底。
这题交G++跑1000MS+,但据说数据乱搞直接平方级暴力可以100MS+……
/* Author : Speedcell Update : 2013-05-30 Version : soppYcell 2.2(a) */ #include <algorithm> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <map> #include <set> #include <list> #include <stack> #include <queue> #include <deque> #include <vector> #include <string> #include <bitset> #include <memory> #include <complex> #include <numeric> #include <stdio.h> #include <stdlib.h> #include <string.h> #include <math.h> #include <time.h> #include <ctype.h> #include <locale.h> using namespace std; #pragma pack(4) #ifndef __CONSTANT__ #define __CONSTANT__ typedef long long LONG; const double pi = acos(-1.0); const int inf = 0x7f7f7f7f; const LONG INF = 0x7f7f7f7f7f7f7f7fll; const int go[8][2] = {{0,1},{0,-1},{1,0},{-1,0},{1,1},{1,-1},{-1,1},{-1,-1}}; #endif // __CONSTANT__ #ifndef __IO__ #define __IO__ inline bool RD(int & a) {return scanf("%d",&a)!=EOF;} inline bool RD(char & a) {return scanf("%c",&a)!=EOF;} inline bool RD(char * a) {return scanf("%s", a)!=EOF;} inline bool RD(double & a) {return scanf("%lf",&a)!=EOF;} inline bool RD(LONG & a) {return scanf("%I64d",&a)!=EOF;} template<class T1> inline bool IN(T1 & a) {return RD(a);} template<class T1,class T2> inline bool IN(T1 & a,T2 & b) {return RD(a)&&RD(b);} template<class T1,class T2,class T3> inline bool IN(T1 & a,T2 & b,T3 & c) {return RD(a)&&RD(b)&&RD(c);} template<class T1,class T2,class T3,class T4> inline bool IN(T1 & a,T2 & b,T3 & c,T4 & d) {return RD(a)&&RD(b)&&RD(c)&&RD(d);} template<class T1,class T2,class T3,class T4,class T5> inline bool IN(T1 & a,T2 & b,T3 & c,T4 & d,T5 & e) {return RD(a)&&RD(b)&&RD(c)&&RD(d)&&RD(e);} template<class T1,class T2,class T3,class T4,class T5,class T6> inline bool IN(T1 & a,T2 & b,T3 & c,T4 & d,T5 & e,T6 & f) {return RD(a)&&RD(b)&&RD(c)&&RD(d)&&RD(e)&&RD(f);} template<class T1,class T2,class T3,class T4,class T5,class T6,class T7> inline bool IN(T1 & a,T2 & b,T3 & c,T4 & d,T5 & e,T6 & f,T7 & g) {return RD(a)&&RD(b)&&RD(c)&&RD(d)&&RD(e)&&RD(f)&&RD(g);} inline void PT(int a) {printf("%d",a);} inline void PT(char a) {printf("%c",a);} inline void PT(char * a) {printf("%s",a);} inline void PT(double a) {printf("%f",a);} inline void PT(LONG a) {printf("%I64d",a);} inline void PT(const char a[]) {printf("%s",a);} template<class T1> inline void OT(T1 a) {PT(a);} template<class T1,class T2> inline void OT(T1 a,T2 b) {PT(a),PT(' '),PT(b);} template<class T1,class T2,class T3> inline void OT(T1 a,T2 b,T3 c) {PT(a),PT(' '),PT(b),PT(' '),PT(c);} template<class T1,class T2,class T3,class T4> inline void OT(T1 a,T2 b,T3 c,T4 d) {PT(a),PT(' '),PT(b),PT(' '),PT(c),PT(' '),PT(d);} template<class T1,class T2,class T3,class T4,class T5> inline void OT(T1 a,T2 b,T3 c,T4 d,T5 e) {PT(a),PT(' '),PT(b),PT(' '),PT(c),PT(' '),PT(d),PT(' '),PT(e);} template<class T1,class T2,class T3,class T4,class T5,class T6> inline void OT(T1 a,T2 b,T3 c,T4 d,T5 e,T6 f) {PT(a),PT(' '),PT(b),PT(' '),PT(c),PT(' '),PT(d),PT(' '),PT(e),PT(' '),PT(f);} template<class T1,class T2,class T3,class T4,class T5,class T6,class T7> inline void OT(T1 a,T2 b,T3 c,T4 d,T5 e,T6 f,T7 g) {PT(a),PT(' '),PT(b),PT(' '),PT(c),PT(' '),PT(d),PT(' '),PT(e),PT(' '),PT(f),PT(' '),PT(g);} template<class T1> inline void OL(T1 a) {PT(a),PT('\n');} template<class T1,class T2> inline void OL(T1 a,T2 b) {PT(a),PT(' '),PT(b),PT('\n');} template<class T1,class T2,class T3> inline void OL(T1 a,T2 b,T3 c) {PT(a),PT(' '),PT(b),PT(' '),PT(c),PT('\n');} template<class T1,class T2,class T3,class T4> inline void OL(T1 a,T2 b,T3 c,T4 d) {PT(a),PT(' '),PT(b),PT(' '),PT(c),PT(' '),PT(d),PT('\n');} template<class T1,class T2,class T3,class T4,class T5> inline void OL(T1 a,T2 b,T3 c,T4 d,T5 e) {PT(a),PT(' '),PT(b),PT(' '),PT(c),PT(' '),PT(d),PT(' '),PT(e),PT('\n');} template<class T1,class T2,class T3,class T4,class T5,class T6> inline void OL(T1 a,T2 b,T3 c,T4 d,T5 e,T6 f) {PT(a),PT(' '),PT(b),PT(' '),PT(c),PT(' '),PT(d),PT(' '),PT(e),PT(' '),PT(f),PT('\n');} template<class T1,class T2,class T3,class T4,class T5,class T6,class T7> inline void OL(T1 a,T2 b,T3 c,T4 d,T5 e,T6 f,T7 g) {PT(a),PT(' '),PT(b),PT(' '),PT(c),PT(' '),PT(d),PT(' '),PT(e),PT(' '),PT(f),PT(' '),PT(g),PT('\n');} #endif // __IO__ #ifndef __MACRO__ #define __MACRO__ #define ML(times) int tcase; IN(tcase); FOR(times,1,tcase) #define FOR(i,a,b) for(int i=int(a),_##i=int(b);i<=_##i;i++) #define DWN(i,b,a) for(int i=int(b),_##i=int(a);_##i<=i;i--) #define ECH(i,u,pre,next) for(int i=int(pre[u]);i!=-1;i=int(next[i])) #define MEM(a,v) memset(a,v,sizeof(a)) #define CLR(a,v) FOR(_i##a,0,sizeof(a)/sizeof(a[0])-1) a[_i##a]=v #define LOOP(a,n) \ FOR(_i##a,0,(n)-1) \ OT(a[_i##a]),OT(_i##a!=__i##a?' ':'\n') #define LOOP2(a,n,m) \ FOR(_i##a,0,(n)-1) FOR(_j##a,0,(m)-1) \ OT(a[_i##a][_j##a]),OT(_j##a!=__j##a?' ':'\n') #define LOOPG(G,n,pre,next) \ FOR(_i##a,0,(n)-1) ECH(_j##a,_i##a,pre,next) \ OL(_i##a,G[_j##a].v,G[_j##a].w) #endif // __MACRO__ #ifndef __BIT__ #define __BIT__ template<class T> inline T lb(T i) {return i&-i;} template<class T> inline T lc(T i) {return i<<1;} template<class T> inline T rc(T i) {return i<<1|1;} template<class T> inline T at(T a,int i) {return a& (T(1)<<i);} template<class T> inline T nt(T a,int i) {return a^ (T(1)<<i);} template<class T> inline T s1(T a,int i) {return a| (T(1)<<i);} template<class T> inline T s0(T a,int i) {return a&~(T(1)<<i);} #endif // __BIT__ #ifndef __DOUBLE__ #define __DOUBLE__ const double eps = 1e-8; inline int cmp(double a,double b=0) {return fabs(a-b)<eps?0:((a-b)>eps?+1:-1);} inline double fmax(double a,double b) {return cmp(a,b)>0?a:b;} inline double fmin(double a,double b) {return cmp(a,b)<0?a:b;} #endif // __DOUBLE__ const int MAXV = 100002; class SegmentTree { private: int L,R,a[MAXV<<2]; #define ls(l,r,i) l,(l+r)>>1,lc(i) #define rs(l,r,i) ((l+r)>>1)+1,r,rc(i) void add(int v,int l,int r,int i) { if(l<=r) a[i]|=v; } void up(int l,int r,int i) { if(l<r) a[i]=a[lc(i)]|a[rc(i)]; } void update(int x,int v,int l,int r,int i) { if(x==l&&r==x) add(v,l,r,i); else { if(x<=(l+r)/2) update(x,v,ls(l,r,i)); else update(x,v,rs(l,r,i)); up(l,r,i); } } int query(int v,int now,int l,int r,int i) { if(l>r) return -1; else if((a[i]|now)<v) return r; else { if(l==r) return -1; else if((a[lc(i)]|now)>=v) return query(v,now,ls(l,r,i)); else return max((l+r)/2,query(v,a[lc(i)]|now,rs(l,r,i))); } } public: void clear(int l,int r) { L=l; R=r; MEM(a,0); } void update(int x,int v) { update(x,v,L,R,1); } int query(int x,int v) { int ans=query(v,0,L,R,1); if(ans<x) return 0; else return ans-x+1; } }st; int n,m,v[MAXV]; int main() { #ifndef ONLINE_JUDGE freopen("A Bit Fun.txt","r",stdin); #else #endif ML(times) { IN(n,m); int ans=0; st.clear(0,n-1); FOR(i,0,n-1) IN(v[i]); DWN(i,n-1,0) { st.update(i,v[i]); ans+=st.query(i,m); } printf("Case #%d: %d\n",times,ans); } return 0; }