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1103: [POI2007] Metropolitan meg time limit:10 Sec Memory limit:162 MBsubmit:2009 solved:1056[Submit] [Status] [Discuss] DescriptionUnder the influence of the tide of economic globalization, the postman, who is accustomed to strolling in the early morning countryside, has also begun to send mail by motorbike. However, she often recalls the past in the country walks the scene. In the past, the countryside was numbered 1. N of n small villages,

();intf=1, x=0; A while(ch'0'|| Ch>'9'){if(ch=='-') f=-1; Ch=GetChar ();} at while(ch>='0'ch'9') x=x*Ten+ch-'0', ch=GetChar (); - returnf*x; - } - intc[maxn*2],n; - voidUpdateintXintV) { while(x2) c[x]+=v,x+=lb (x);} - intQueryintx) {intq=0; while(x) q+=c[x],x-=lb (x);returnq;} in intMain () { -N=read (); Inc (I,1, N-1){intA=read (), b=read (); PE (A, b);} toDfs1); Inc (I,1, n) update (L[i],1), Update (r[i],-1); + intM=read ();Charopt[3]; -Inc (I,1, n+m-1){ thescanf"%s", opt);

[eid]=a; NEXT[EID]=G[B]; g[b]=eid++;}voidDfs () {s[++sp]=1; while(sp) {intu=s[sp--]; if(!L[u]) {L[u]=++DFN; s[++sp]=T; for(intI=g[u];~i;i=next[i])if(v[i]!=Fa[u]) {Fa[v[i]]=u; s[++sp]=V[i]; } } Elser[u]=++DFN; }}intMain () {scanf ("%d",N); Memset (g,-1,sizeof(g)); Bit.init (n*2); for(intI=1, a,b;i) {scanf ("%d%d",a,b); Addedge (A, b); } dfs (); for(intI=2; i) {Bit.add (l[i],1); Bit.add (R[i],-1); } Charop[Ten]; scanf ("%d", m); m+= (n1); for(intI=1, a,b;i) {scanf (

Tree links are separated by water, and can be modified at a single point. You can use a tree array. # Include 1103 poi2007 metropolitan Meg

Preprocessing each point to the root node of the soil, inserted into a tree array, and then each modification will only affect the nodes in the subtree, thus equivalent to the interval modification, point query.#include Dfs sequence "tree array" bzoj1103 [POI2007] Metropolitan Meg

1103: [POI2007] Metropolitan Meg Time Limit:10 Sec Memory limit:162 MBsubmit:2324 solved:1231[Submit] [Status] [Discuss] Description Under the influence of the economic globalization, the postman Blue Mary, who was accustomed to strolling in the early morning Country Road, began to deliver the mail on her motorbike.However, she often recalls her previous stroll in the country. In the old days, the countryside numbered 1 sequentially. n small villages,

1103: [POI2007] Metropolitan meg time limit:10 Sec Memory limit:162 MBsubmit:2859 solved:1505[Submit] [Status] [Discuss] DescriptionUnder the influence of the tide of economic globalization, the postman, who is accustomed to strolling in the early morning countryside, has also begun to send mail by motorbike.However, she often recalls the past in the country walks the scene. In the past, the countryside was numbered 1. n small villages, some

1103: [POI2007] Metropolitan Meg Time Limit:10 SecMemory limit:162 MBDescription Under the influence of the tide of economic globalization, the postman, who is accustomed to strolling in the early morning countryside, has also begun to send mail by motorbike. However, she often recalls the past in the country walks the scene. In the past, the countryside was numbered 1. N of n small villages, some villages have two-way dirt roads. From each villag

]:=Time ; -e:=Head[u]; - whileE0 Do + begin -v:=Vet[e]; + ifflag[v]=0 ThenDfs (v); Ae:=Next[e]; at End; -Inc (time); A[time]:=u; c[u]:=Time ; - End; - - functionlowbit (x:longint): Longint; - begin inExit (x and(-x)); - End; to + functionsum (x:longint): Longint; - begin thesum:=0; * whileX>0 Do $ beginPanax Notoginsengsum:=sum+T[x]; -x:=x-lowbit (x); the End; + End; A the procedureUpdate (x,y:longint); + begin - whilex1 Do $ begin $t[x]:=t[x]+y; -x:=x+lowbit (x); - End; the

(Request_code_camera, Monhanlderresultcallback);With configuration galleryfinal. Openedit(Request_code_camera, Functionconfig, Monhanlderresultcallback);Add a callback returned by an operation privatenew GalleryFinal.OnHanlderResultCallback() { @Override publicvoidonHanlderSuccess(int reqeustCode, Listifnull) { mPhotoList.addAll(resultList); mChoosePhotoListAdapter.notifyDataSetChanged(); } }Then it can be used, it seems to need to c

; - inthead[n],to[n],next[n],cnt; to voidAddintXinty) { +To[++cnt]=y; NEXT[CNT]=HEAD[X]; head[x]=CNT; - } the /********************edge**********************/ * intn,m,dfs_clock,t[n*2],l[n],r[n]; $InlineintLowbit (intx) {returnx (-x);}Panax Notoginseng voidUpdateintXintval) { - for(inti=x;i2; i+=lowbit (i)) thet[i]+=Val; + } A intSumintx) { the inttemp=0; + //for (int i=1;i - for(intI=x;i;i-=lowbit (i)) temp+=T[i]; $ returntemp; $ } - intfa[n],st[n],top=0; - voidDfs () { thest[

http://www.lydsy.com/JudgeOnline/problem.php?id=1103Test instructions: An N-node tree (1 #include 　　 The DFS sequence should be ...Obviously, every time you change an edge, it affects all subtrees of a node with a greater depth, and asking is asking 1-the right value of the current point and the line.So let's maintain the DFS sequence and prefix and"Bzoj" 1103: [POI2007] Metropolitan Meg

Title DescriptionSolutionThe exam was not carefully thought, the results only took the violent points ...is actually a DFS sequence + tree-like array.Let's turn it into a sequence with DFS and record their \ (siz\).So every time we have one edge, we have an effect on the subtree.It's good to maintain it in a tree array.Code#include [Poi2007]meg-megalopolis (tree array, DFS order)

The main idea: to give a tree, the beginning of every two points are connected by dirt road, but there will be some dirt road gradually into the road, asked each time from point 1 to the number of points K how many dirt road.Ideas: POI is not difficult problem, in fact, each point to 1 of the number of dirt road is the depth of this point, in the dirt road into the road, the point and the subtree of all the nodes of the depth of 1, sub-tree modification is very basic, you can use the DFS sequenc

1103: [poi2007] megtime limit: 10 sec memory limit: 162 MB Submit: 1145 solved: 569 [Submit] [Status] Description Under the influence of the wave of economic globalization, the postman Blue Mary, who is used to walking in the village path in the

Description In the old days, the countryside numbered 1 sequentially. n small villages, some villages have a two-way dirt road. From every village there is exactly one path to the village 1 (that is, Fort bitcoin). Also, for each village, its path

#include#include#include#includeSet>#definell Long Long#defineMM 264005#defineMN 500005using namespacestd;structmeg{intKi,pos,lf,rf,val,aps;} PX[MN];structnode{intG,x,y;} B[MM];Setint>Se[mm];ll t[mm],ans[mm];intc[mm][4],las[mm],pre[mm],col[mm];intMq,n,m,pxin;inlineintRead () {intn=0, f=1;CharC=GetChar (); while(c'0'|| C>'9') {if(c=='-') f=-1; C=GetChar ();} while(c>='0' c'9') {n=n*Ten+c-'0'; C=GetChar ();} returnNF;} InlinevoidGetadd (intXintZ) { for(x+=mq;x;x>>=1) t[x]+=Z;} inline ll Getsum

Operating environment: Eclipse | | MyEclipsePackage socaket;These classes are all placed under the Socaket package.This is the server-side message sending classImport java.io.IOException;Import Java.io.PrintWriter;Import Java.net.Socket;Import Java.util.Scanner;public class Serverchatsend extends Thread {Server-Side Send threadSocket SSK = null;constructor functionServerchatsend (Socket SK) {This.ssk = SK;}public void Serverchatsend () {try {Socket SK = Ssk.accept ();PrintWriter pw = new PrintWr

= elementName; if ([elementName isinclutostring: @ "message"]) {Recorce * meg = [[Recorce alloc] init]; [self. megArray addObject: meg] ;}// locate content (node)-(void) parser :( NSXMLParser *) parser foundCharacters :( NSString *) string {Recorce * meg = [self. megArray lastObject]; // Add content through the node [meg