at t independence mo

Topic Portal1 /*2 Test Instructions: The number of occurrences of x within the interval3 MO Team algorithm: With a CNT record x number of times you can4 and a binary search method.5 */6#include 7#include 8#include 9#include Ten using namespacestd; One A Const intMAXN = 1e5 +Ten; - Const intINF =0x3f3f3f3f; - structData the { - intB, L, r, X; - intID; - }DATA[MAXN]; + intA[MAXN]; - intCNT[MAXN]; + intANS[MAXN]; A intN, Q; at - BOOLCMP (data x

#include using namespace STD;TemplateintN>classbitset{ Public: BitSet () {Set(); }void Set() { for(inti =0; i 0; } }void Set(intN) {vtr[(n1) / +] |= (0x11) % +)); }voidPrintf () { for(inti = N; i >0; i--) {cout1)/ +]>> ((I-1)% +)) 0x1) ==0?"0":"1"); } }BOOLTestintN) {if(Vtr[(n1) / +] >> (n1) % +) 0x1)return true;return false; }Private:enum{_n_= (n1) / (sizeof(size_t) *8)};enum{nm=_n_+1};intVTR[NM];};intMain () {bitset344> bt; Bt.Set( +);//So a simple bitset is done to find out whether

]; - intp=0;Wuyi for(intj=i+1; j) the { - if(JGT;REG[I][P].RR) p++; Wupre[j]=REG[I][P].GCD; -pre[j]+=pre[j-1]; About } $ for(intj=1; j) - if(I>=que[j]. L iQue[j]. R) -QUE[J].ANS+=PRE[QUE[J]. r]-pre[i-1]; - } A for(intI=1; i"%d\n", Que[i].ans); + } the intMain () - { $ //freopen ("Input.txt", "R", stdin); the intT, L, R; theCin>>T; the while(t--) the { -scanf"%d", n); in for(intI=1; i"%d", seq[i]

]; }}q[n];structcmeow{intP,v,last;} Cq[n];intP,tim, ans[n];intC[m], now;intL=1, r=0, cur=0; inlinevoidAddintx) {now+= (++c[x]) = =1;} InlinevoidDelintx) {now-= (--c[x]) = =0;} InlinevoidChaintPintv) { if(Lr) Add (v), Del (A[p]); A[P]=v;}voidModui () { for(intI=1; i){ while(Cur, Cha (cq[cur].p, CQ[CUR].V); while(Cur>q[i].tim) Cha (cq[cur].p, cq[cur].last), cur--; while(R, add (A[r]); while(RGT;Q[I].R) del (A[r]), r--; while(L; while(LGT;Q[I].L) l--, add (A[l]); ans[Q[i].id]=Now ;

Tags: des style blog HTTP color AR for strong sp Cause: Suppose you want to mount high-pollution projects, such as PX, where you live. Encourage others to go to the streets for violent demonstrations and confrontation with military and police personnel. They will stay at home to support such behavior. Such a person is really wretched! Because he often changes his post without permission, he will keep it here. Really boring. actually blocking others' IDS as others "do not fight back and swe

Category of Mo Function Points 1. Enable (AE) for the entire process, and mount mapcontrol event (simplified to H below), for example, dynamic coordinate display 2. Instantaneous enable (IE) of the Action. mapcontrol event is not attached, but attribute of the read period, such as layer settings. 2. The instantaneous enable (IE) action is mounted to the mapcontrol event and the execution condition reaches the trigger condition in a function component,

);}Const intN = 1e7+7;intVis[n],g[n],p[n],a[n],val[n];voidinit () {rep0 (I,2, N) { if(Vis[i] = =0) {p[++p[0]] =i; G[i]=1; A[i]=1; Val[i]=i; } for(intj =1;p [J]*i ) {vis[i*P[J]] =1; if(I%p[j] = =0) {A[i*P[J]] = a[i]+1;//the power exponent of the minimum factorization;VAL[I*P[J]] = val[i]*P[j]; intTMP = i/Val[i]; if(TMP = =1) G[i*p[j]] =1;//there's only one quality factor . ElseG[i*P[J]] = (a[tmp] = = A[i*p[j]]?-g[tmp]:0); Break; } A[p[j]*i] =1;//indicates the number

+ R >>1; + if(L mid) Solve (L, mid); -Sort (A + L, a + R +1, CMP); $Times + +; $ for(inti = l; I ) - if(A[i].id mid) Add (A[i].z, num[a[i].id]); - ElseRank[a[i].id] + =query (a[i].z); theSort (A + L, a + R +1); - if(Mid +11, R);Wuyi } the intMain () { -Freopen ("3262.in","R", stdin); WuFreopen ("3262.out","W", stdout); -scanf"%d%d",n,k); About for(inti =1; I ) $scanf"%d%d%d", a[i].x, AMP;A[I].Y, a[i].z); -Sort (A +1, A + n +1); - for(inti =1; I A[i]; -

} - Elsemiu[i*prime[j]]=-Miu[i]; to } + - } the } * Long LongCalcintNintm) $ {Panax Notoginseng Long Longret=0; - if(n>m) swap (n,m); the intnext=0; + for(intI=1; i1) A { theNext=min (n/(n/i), m/(m/i)); +ret+= (Long Long) (sum[next]-sum[i-1]) * (n/i) * (m/i); - } $ returnret; $ } - intMain () - { the Mobius (); - for(intI=1; i)Wuyisum[i]=sum[i-1]+Miu[i]; the inta,b,c,d,k,t; -scanf"%d",t); Wu while(t--) -

and 3D hubs. From the initial concept to the final finished product, the whole process took nearly 4 months.In an interview, Akhtar explained how the costume was created. My aim is to create a garment that can only be made out of 3D printing in any other way, so it comes with many complex internal honeycomb structures. Its shoulder is a directly printed interlocking structure, while the chest armor is left with ample internal space for the fiber to pass through. I was going to make it bigger an

[prime[j]*i] =true; A if(i%Prime[j]) { atMU[I*PRIME[J]] =-Mu[i]; -F[I*PRIME[J]] =-f[i]+Mu[i]; -}Else{ - /* - in itself I already contain primes prime[j] - If there are at least 3 factor prime[j] in the IF, then at the end of the day, the current value is divided by a factor of at least in contains two prime[j], then it must be 0. - if there are only 2 prime[j], then there are at least two factors in addition to the other cases divided by prime[j]. to , then mu[] must

[i]%MoD; - Break;Wuyi } theh[pri[j]*i]=h[pri[j]]*h[i]%MoD; -J + +; Wu } - } AboutFr (I,1, MAXP) $F[i]= (f[i-1]+h[i])%MoD; - } - - ll Sum (ll X,ll y) A { + return((x+1) *x/2%MOD) * ((y+1) *y/2%MOD)%MoD; the } - $ the ll Calc (ll X,ll y) the { the if(x>y) Swap (x, y); thell res=0, i=1, POS; - while(ix) { inPos=min (x/(x/i), y/(y/i)); theRes= (Res+sum (x/i,y/i) * (f[pos]-f[i-1])%mod)%MoD; thei=pos+1; About } the return(res+mod)%MoD; the } the + voidWork () - { the

Description god Ben yy after abusing the number theory to give silly xkac out a question given N, M, ask 1 KAC this silly x Must not, so to you to consult ... multiple sets of inputs The surface is still very kind, and the difference before is that all prime numbers then bashi the Möbius function and ask for a strange prefix andThe time complexity here is quite magical .1#include 2#include 3#include 4 Const intMAXN =10000010, Maxp =664580;5 intVIS[MAXN],PRIME[MAXP

);//freopen("2.txt","W", stdout); Init (); scanf"%d", t); while(t--) {scanf ("%d%d",n,m); k=sqrt(n*.0); for(i=1; i"%d", a[i]); } for(i=0;im; i++) {scanf ("%d%d", AMP;FEI[I].L,AMP;FEI[I].R); Fei[i].id=i; Fei[i].POS=fei[i].l/k; }Sort(fei,fei+m, CMP); L=1; r=1; res=1; memset (Ha,0, sizeof (HA)); ha[a[1]]=1; for(i=0;im; i++) { while(R*(r-l+1)%mod*INV[Ha[a[r]]%mod; } while(RGT;FEI[I].R) {Res=res*ha[A[r]]%mod*INV[r-l+1]%mod; ha[a[r]]--;

Break; at}Else{ -MU[I*PRIME[J]] =-Mu[i]; - } - } - } - } in - intMain () to { + //freopen ("a.in", "R", stdin); - GETMU (MAXN); the intT, N; *scanf"%d", T); $ while(t--)Panax Notoginseng { -scanf"%d", n); thell ans =0; + for(intI=1; I){ All tmp1 = (ll) (n/i) * (n/i) * (n/i) *mu[i];//a situation where a 0 doesn't exist in the coordinates. thell tmp2 = (ll)3* (n/i) * (n/i) *mu[i];//there's a zero in the coordinates. +ll Tmp3 = (ll

(ch[k][0],x); the Else returnPp+c[k]+findrank (ch[k][1],x); - } $}t[200020]; the intans[100010]; the intn,m; theInlineintLowbit (intx) {returnx-x;} the voidAddintKintx) - { in while(km) the { the T[k].add (t[k].root,x); Aboutk+=Lowbit (k); the } the } the intQueryintKintx) + { - intret=0; the while(k)Bayi { theret+=T[k].findrank (t[k].root,x); thek-=Lowbit (k); - } - returnret; the } the intsum[100010]; the intCSC () the { -Freopen ("in.in","R", std

.XAMP;AMP;A1.Y==A2.Y)Return a1.zif (a1.x==a2.x)Return a1.yReturn a1.x}void Update (int A1){SZ1[A1]=SZ1[LR[A1]]+SZ1[RR[A1]]+W[A1];}void Zuo (int a1){int T=LR[A1];LR[A1]=RR[T];RR[T]=A1;Update (A1);Update (t);a1=t;}void You (int a1){int T=RR[A1];RR[A1]=LR[T];LR[T]=A1;Update (A1);Update (t);a1=t;}void JIAA (int a1,int A2){if (a1==0){sz++;A1=sz;V[A1]=A2;Sz1[a1]=1;W[a1]=1;Rou[a1]=rand ();Return}sz1[a1]++;if (V[A1]==A2){w[a1]++;Return}if (V[A1]GT;A2){JIAA (LR[A1],A2);if (ROU[LR[A1]]Zuo (A1);}Else{JIAA

format: The First behavior n,k (1 The following n lines, three integers per line, si, CI, mi (1 output format: contains n rows, each of which represents a rating of 0 ... N-1 the number of flowers per level. Practice: Three-dimensional, is said to be able to write with a tree set of trees (this later try to write a pitch), and then this problem is the traditional CDQ Division + sorting + Tree array, set the flower of three properties for X, Y, Z, we will spend the first keyword, the second key

See for a long time also did not understand Möbius is what, first put a piece of code it, as if to ask the value of a function of the inverse of the mo.int mobi (int n) { int m=1; for (int i=2;i*iMöbius inversion study "the inverse of the Mo-Wu-yan"

Article XiaolongFrom Public Account: product prototype craftsman You can use deformation to enrich the top menu 1. Pull a rectangle with a width of 375 and a height of 60 at the position of X: 0, Y: 202. The default size of a left-facing icon is X: 26, Y: 383. The default size of a right icon. The position is the same as that of the left icon to hide it. 1. Pull a global gesture first. 2. Add status 23. Hide to left. Show to right. 4. Set the rounded corner of the rec