666 socks

()) = ='-') flag=1; Else if(ch>='0'ch'9') res=ch-'0'; while((Ch=getchar ()) >='0'ch'9') res=res*Ten+ (ch-'0'); returnflag?-res:res;}voidOut (inta) {if(a0) {Putchar ('-'); a=-A;} if(a>=Ten) Out (A/Ten); Putchar (A%Ten+'0');}Const intn=50005;//Code begin ...structquery{intL, R, id;} Node[n]; ll GCD (ll A, ll b) {returnb==0? A:GCD (b,a%b);}structans{LL A, B; voidReduce () {LL d=gcd (A, b); a/=d; b/=D;}} Ans[n];intA[n], num[n], N, M, unit;BOOLComp (query A, query B) {if(A.l/unit!=b.l/unit)returna.

Offline practices: Mo team.Offline processing of queries, the left end of the block, in the same piece of the right end of the order from small to large.Then, for the previous query [L,r] and today's inquiry [l ', R '] constantly adjust the range of intervals.How to adjust it? For the information contained in [L,r], we can adjust the time of O (1) to [L,r+1],[l,r-1],[l+1,r],[l-1,r].The most authentic Mo team seems to be the point set Manhattan distance plus the smallest spanning tree, very troub

(; t) -Up (t+1,1); in for(; t>da[i].r;t--) theUp (t,-1); the for(; h) AboutUp (h,-1); the for(; h>da[i].l;h--) theUp (H-1,1); the if(da[i].l==DA[I].R) + { -Da[i].a=0; theda[i].b=1;Bayi Continue; the } thell len=da[i].r-da[i].l+1; -da[i].a=s-Len; -da[i].b=len* (len-1); thell k=gcd (da[i].a,da[i].b); theDa[i].a/=K; theDa[i].b/=K; the } -Sort (da+1, da+m+1, CMP1); the for(intI=1; i) theprintf"%lld/%lld\n", da[i].a,da[i]

; + } - the intMain () * { $ intN, M;Panax Notoginseng while(~SCANF ("%d%d", n, m)) { - for(inti =1; I ) { thescanf"%d", A +i); + } A for(inti =1; I ) { thescanf"%d%d", AMP;Q[I].L, Q[I].R); +Q[i].id =i; - } $U =sqrt (n); $Sort (q +1, q + M +1, CMP); - intL =1, R =0; -LL temp =0; thememset (cont,0,sizeof(cont)); - for(inti =1; I ) {Wuyi while(L Q[I].L) { theTemp-=f (Cont[a[l]]); -cont[a[l]]--; WuTemp + =f (Cont[a[l]]); -l++

;) { if(j1, Num[c[j]]--, j + +; Elseans[i]+=num[c[j-1]],num[c[j-1]]++,j--; } for(intj=que[i-1].r,to=que[i].r;j!=to ;) { if(j1]],num[c[j+1]]++,j++; Elseans[i]-=num[c[j]]-1, Num[c[j]]--, j--; } Ans[i]+=ans[i-1]; } intDIV=GCD (Ans[i], (que[i].r-que[i].l+1) * (QUE[I].R-QUE[I].L)/2); if(que[i].r==que[i].l| | ans[i]==0) ans1[que[i].num][0]=0, ans1[que[i].num][1]=1; Elseans1[que[i].num][0]=ans[i]/div,ans1[que[i].num][1]= (que[i].r-que[i].l+1) * (QUE

MSFT announced azure .) Will it scale? It's conventional wisdom that relational databases aren't as scalable as the schemaless entity-Attribute-value (EAV) tables used by SSDS, original SDS, Amazon simpledb, and Google's bigtable and the App Engine's data store. if the SSDS team was leveling with us when they attributed the choice of the Eav (authority-container-entity) data model to simplifying data access (not scalability ), it cocould turn out that conventional wisdom was wrong in this case.

ans;structn{intL,r,bh;ll A, B;} Q[MAXN];intRd () {intret=0;CharCh=GetChar (); while(ch'0'|| Ch>'9') ch=GetChar (); while(ch>='0'ch'9') ret=ret*Ten+ch-'0', ch=GetChar (); returnret;}BOOLCMP (N x,n y) {returnblk[x.l]==blk[y.l]?x.rBLK[Y.L];}BOOLCMP2 (N x,n y) {returnx.bhy.bh;} ll C (ll x) {returnx* (x1)/2;} ll GCD (ll A,ll b) {returna%b==0? B:GCD (b,a%b);}voidPopintx) {cnt[c[x]]--;ans-=cnt[c[x]];}//Note OrdervoidPushintx) {ans+=cnt[c[x]];cnt[c[x]]++;}intMain () {n=rd (); M=rd (); k=sqrt (n); fo

. inputFirst line: An integer nthat represents the length of the sequence.Second line:n integers A1,A2,A3, ..., an(1 ), The number and number are separated by a space.Third line: An integer q (that indicates the number of queries.Next Q line, two integers per line i,J(1 ), indicates the interval to ask.OutputFor each query, output the number of different numbers within this interval.Sample Input51 1 2) 1 331 52 43 5Sample Output323No special.#include#include#include#includeusing namespace std;#d

plus 1temp+= (Long Long) num[a[r]]*Num[a[r]]; } while(l>M[i]. L) {L--; Temp-=(Long Long) num[a[l]]*Num[a[l]]; NUM[A[L]]++;//A[l] The number plus 1temp+= (Long Long) num[a[l]]*Num[a[l]]; } while(r>M[i]. R) {Temp-=(Long Long) num[a[r]]*Num[a[r]]; NUM[A[R]]--;//A[l] The number minus 1temp+= (Long Long) num[a[r]]*Num[a[r]]; R--; } while(lM[i]. L) {Temp-=(Long Long) num[a[l]]*Num[a[l]]; NUM[A[L]]--;//A[r] The number minus 1temp+= (Long Long) num[a[l]]*Num[a[l]]; L++; } ans[m[

1#include 2#include 3#include 4#include 5#include 6 using namespacestd;7 #defineMAXN 500058typedefLong Longll;9ll ans=0, ANS[MAXN],BNS[MAXN];Ten intn,m,sum[maxn],a[maxn],block[maxn],temp; One structtsegment{ A intId,l,r; - }QS[MAXN]; - voidprepare () { thetemp=sqrt (n); - for(intI=1; i1)/temp+1; - } - BOOLComp (tsegment x,tsegment y) { + if(block[x.l]==BLOCK[Y.L]) { - if(block[x.l]%2==1)returnx.rY.R; + Else returnX.r>Y.R; A } at returnblock[x.l]BLOCK[Y.L]; -

1#include 2#include 3#include 4 5 using namespacestd;6 7Unsignedintr,sum,p,q;8Unsignedintst[50010][2];9 Ten unsigned gcd (unsigned a,unsigned b) One { A returnB?GCD (b,a%b): A; - } - /*Hash Linked list*/ the Const intHash=50007; - intHead[hash],next1[hash]; - intHASH1 (unsignedints) - { + returns%HASH; - } + intInsertints) A { at intH=HASH1 (st[s][0]); - intu=Head[h]; - while(u)//The end of the list is always 0, so s cannot be 0 -u=Next1[u]; -NEXT1[S]=HEAD[H];//the original

Test instructions: Chinese question.Analysis: The very famous MO team algorithm, the first expression of this probability, should be the molecule: C (x1, 2) + C (x2, 2) + C (X3, 2) + ... + c (xn, 2) Denominator: C (n, 2), and then the expression of fractions, [X1 (x1-1) +x2 (x2-1) +...+XN (xn-1)]/(n (n-1)) then the simplification is obtained (Sigma (XI*XI)-N)/(n (n-1)), then the operation of each interval, offline, so that the sequence is divided into sqrt (n) block, and then with two pointers,

1. Create tunnel on the server using SSHSSH 10080 -f-c-q-n [email protected]112.34. 113.35 -D: Let SSH on the specified port (1025-65536-----n:tells SSH That no command would be sent once the tunnel is up# the last specified is the SSH user and the currently established tunnel server IPRun in the background after execution, you can view the process through PS2. Configure SOCKS5 proxy locally via puttyOpen Putty,1) Session, fill in the Host Name is just the server IP, port is 22 (note that this

[I].R); theb[i].k=b[i].l/sq; -B[i].id=i;Wuyi } theSort (b +1, b+m+1, CMP1); -memset (SUM,0,sizeof(sum)); Wu intL=1, r=1; sum[a[1]]=1; ans=0; - for(intI=1; i) About { $ while(L1); l++;} - while(LGT;B[I].L) {Revise (L-1,1); l--;} - while(R1,1); r++;} - while(RGT;B[I].R) {Revise (r,-1); r--;} AB[i].fz=ans;b[i].fm= (LL) (b[i].r-b[i].l+1)) * ((LL) (B[I].R-B[I].L))/2; +Gcd=GETGCD (b[i].fz,b[i].fm); theB[i].fz/=gcd;b[i].fm/=gcd; - if(b[i].fz=

The main topic: given n points, each point has a color, m inquiry, each asked a range randomly selected two points the same probability of the same numberOtz-mo Team algorithm ...The specifics of this, whether it's a block or a Manhattan minimum spanning tree, I'm not going to talk about it.The practice of this problem is to record a CNT array representing how many of each color in the current intervalAdd a point with a color of x on ans+=cnt[x] and cnt[x]++Reduce the point of a color x to cnt[x

3Proxy is developed by the Russians (how are the Russians, front nginx), the homepage in www.3proxy.ru. 3Proxy supports a variety of platforms such as Windows and Linux. Download the installation directly under Windows, Linux needs to compile itself below. Although it is compiled and installed, it is effortless. download, compile, install: wget http://www.3proxy.ru/current/3proxy-0.7-devel.tgz Tar zxf 3proxy-0.7-devel.tgz CD 3proxy-0.7 Make–f Makefile.linux Make–f Makefile.linux Install

Topic Link: Click to open the linkFirst, the query is processed into Manhattan's smallest spanning tree.Then you can run violently on the tree.Can use MO team situation should be for inquiry [L,r], [l ', R '] cost must be ABS (L-L ') + ABS (R-R ')#include 2038: [2009 countries Training team] small Z socks (hose) MO Team Algorithm

https://github.com/sensepost/reGeorgExamples of use:python reGeorgSocksProxy.py -p 8080 -u http://upload.sensepost.net:8080/tunnel/tunnel.jspStep 1. Upload a script (aspx|ashx|jsp|php) to the Web server. Step 2. Execute the regeorgsocksproxy.py script to connect the shell listening port. * * If you use tools such as nmap need to use Proxychains. Step 3. Hack the planetThis article from "Lonely Moon Cold City" blog, declined reprint!Using Webshell to build so