nvs 285

codeforces Round #285 Div1 A and Div2 Cproblemgive a figure g, make sure G is a forest (pit!) ）。 Figure G contains n points, giving two attributes for each point: degrees (degree), XOR, and (sum). The degree indicates how many points the point is connected to, and the XOR of the number of points to which it is connected (the point number starts at 0, and if the degree is 0, the XOR is 0). Requires the original image, the number of output edges and the

Topic Portal1 /*2 Test Instructions: Give a series of name changes, ask what the last name became3 string Processing: Each time the input to the previous name of the printing, if not, it is the initial, Pos[m] array record the initial position4 the initial POS is added to each update, so the initial name is guaranteed to be updated, which is the only place to think about:)5 */6#include 7#include 8#include 9#include Ten#include One#include string> A#include -#include - using namespacestd; the

Topic Portal1 /*2 Test instructions: gives a non-circular graph, the degree of each point and the xor of the adjacent points (a^b^c^ ...)3 graph theory/bit operations: In fact, the problem is very simple. Similar topological sort, first in the degree of 1 first-in pairs, each one less than one degree4 the key is to update the XOR, the essence: a ^ b = c, a ^ c = b, b ^ c = A;5 */6#include 7#include 8#include 9#include Ten#include One using namespacestd; A - Const intMAXN = 7e4 +Ten; - Const in

Directory1. The basic use of jquery extendThrough 285 lines of source jquery.extend = JQuery.fn.extend = function() { , the Extend method is either a static method that hangs directly into JQuery, Either the attached to FN is actually the instance method on the prototype (referring to the assignment operation of 283 rows). The Extend method can pass an object, similar to a plug-in, or can be copied by multiple objects. Next focus on extend copy and sh

/** 285. Inorder successor in BST * 2016-6-27 by Mingyang * Online Some code is too complex, my most simple and clear, is nothing more than a DFS, with a stack to do just fine, encountered after p * with a flag ma RK a bit, then you can return to the next inorder. * Note: Judging is!stack.empty () and not another method*/ PublicTreeNode inordersuccessor (TreeNode root, TreeNode p) {if(root==NULL) return NULL; StackNewStack(); TreeNod

concept: scope of action. Variables are classified into global variables and local variables based on their defined ranges. Global variables can be used by all scripts. The variable defined in the function is called a local variable, which is only valid in the function. If the global and local variables use the same variable name Local variables overwrite global variables. Code example: // Define the global variable test VaR test = "global variables "; // Define the function checkscope Function

Codeforces #285 solution report A. B .C A-Contest Question. Just calculate the scores of the two people based on the requirements. The Code is as follows: # Include # Include # Include # Include # Include # Include # Include # Include Using namespace std; # define LL _ int64 # define pi acos (-1.0) const int mod = 100000000; const int INF = 0x3f3f3f; const double e

NYOJ 285 search for clone (map + count) Description A town in the United States was recently attacked by aliens. Some residents were taken away and cloned. Now, scientists have extracted the DNA of some people in the town. Please find out the number of DNA with the same number of clones, for example, the following nine Sequences AAAAAAACACACGTTTTGACACACGTTTTGACACACACACACTCCCCCTCCCCCHere, TCCCCC and GTTTTG have two identical individuals, four for ACAC

,sizeof(c)); for(i=1; i1); for(i=1; i) {scanf ("%d",x); X++; P2[i]= Getsum (X-1); Modify (x,-1); } memset (P3,0,sizeof(p3)); for(i=n;i>=1; i--) {P3[i]+ = p1[i]+P2[i]; if(P3[i] >= (n-i+1) ) {P3[i]= p3[i]-(n-i+1); if(I! =1) p3[i-1]++; }} memset (C,0,sizeof(c)); for(i=1; i1);//For (i=1;i//cout//cout for(i=1; i) { intLow =1, high =N; while(Low High ) { intMid = (Low+high)/2; if(Getsum (mid-1) >P3[i]) high= mid-1; Else if(Getsum (mid-1) = = P3[i] getsum

idea: Similar to a topological sorting problem, according to the degree of 1 points, it is connected to the number of edges is different or and this rule, directly to the topological sort. Each time the resulting node is reduced by one, and xor the node he or she is connected to.Code:#include Codeforces Round #285 (Div. 2) C-misha and Forest

Title Link: Http://codeforces.com/contest/501/problem/CTest instructions: Gives the sum of the degrees of each vertex of the graph and the XOR values of the adjacent vertices, and the number of edges and edges of the graph:Analysis: Find the point of a degree, the corresponding XOR value is the adjacent vertex (Code:#include Codeforces #285 Div.2 C. Misha and Forest

queue is empty.1#include 2 3 Const intMAXN = (1 -) +Ten;4 5 intDEGREE[MAXN], XORSUM[MAXN];6 intedge[maxn][2], CNT =0;7 intQ[MAXN], head =0, tail =0;8 intMain ()9 {Ten intN; Onescanf"%d", n); A for(inti =0; I ) - { -scanf"%d%d", degree[i], xorsum[i]); the if(Degree[i] = =1) q[tail++] =i; - } - - while(Head tail) + { - intt = q[head++]; + if(Degree[t] = =0)Continue; ADEGREE[T] =0; at intNeighbor =Xorsum[t]; -edge[cnt][0] = t; edge[cnt++][

Title Descriptionhttp://acm.nyist.net/JudgeOnline/problem.php?pid=285A small town in the United States was recently attacked by aliens and some residents were taken away and cloned, and now scientists have extracted DNA from some people in the town, identifying the number of DNA with the same number of clones, such as the following 9 sequences AaaaaaAcacacGttttgAcacacGttttgAcacacAcacacTcccccTcccccWhere TCCCCC,GTTTTG have two identical individuals, ACACAC have four, aaaaaa is one,

the graph.Next print m lines, each containing, distinct numbers, a and b (0?≤? A? ≤? n?-? 1, 0?≤? b? ≤? n?-? 1), corresponding to edge (a,? b).Edges can printed in any order; Vertices of the edge can also is printed in any order.Sample InputInput32 31 01 0Output21 02 0Input21 11 0Output10 1HintThe XOR sum of numbers is the result of bitwise adding numbers modulo 2. This operation exists in many modern programming languages. For example, in languages C + +, Java and Python it is represented a

Control Panel --> management tools --> servicesThe Disabled services include:Homegroup listener)Homegroup provider (same as above)IP helper)Media Center extender ServicePrint Spooler (do not disable printers)Program compatibility Assistant Service

Test instructions: Give two permutations, find out the rank of each arrangement in the whole arrangement, add, modulo n! (the total number of rows) to obtain a number k, the arrangement of the row behavior K. Solution: First of all to obtain a

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filters, Pins, to do it, are not as simple as I had imagined.At least, I am currently in the kmdod of the transformation process, encountered a series of problems, and then the table.About WinDbg Debugging:WinDbg is a good debugging tool, and there is no comment on this assertion.But my concept still stays on the "shotgun" of com 115200 bps, so the "slow" response to WinDbg is always very impatient.Currently, my configuration is, host Win7 Test Mode, build 7601, slave Win8.1 Pro build 9600.I di