cb5 311

/** 311. Sparse Matrix multiplication * 2016-7-4 by Mingyang * This topic is really the skull into the excrement will be used hashmap do .... */ Public int[] Multiply (int[] A,int[] B) {intm = a.length, n = a[0].length, NB = b[0].length; int[] C =New int[M][NB]; for(inti = 0; I ) { for(intk = 0; K ) { if(a[i][k]! = 0) { for(intj = 0; J ) { if(b[k][j]! = 0) C[I][J] + = a[i][

Codeforces Round #311 (Div. 2) E. Ann and Half-Palindrome (DP + dictionary tree) Question address: PortalUse dp to obtain all the required semi-return strings and mark them. Then construct the dictionary tree. Then, the dfs finds the Subtrees and Subtrees of all nodes, and then finds the k th node.The Code is as follows: #include #include #include #include #include #include #include #includ

?Repository. Products.count ():Repository. Products.where (E = E.category = =category). Count ()},Currentcategory = Category};return View (model);}Public Filecontentresult GetImage (int productId) {Product prod = repository. Products. FirstOrDefault (p = P.productid = = ProductID);if (prod! = null) {Return File (prod. ImageData, prod. Imagemimetype);} else {return null;}}}}This method tries to find a product, matches the ID specified by the parameter. The Filecontentresult class isUsed as the re

; } } } for(intj = 0; J ) { for(inti = 0; i ) { if(b[i][j]! = 0) { int[] tmp =New int[B.length]; for(intk = 0; K ) {Tmp[k]=B[k][j]; } colinb.put (J, TMP); Break; } } } int[] res =New int[A.length] [B[0].length]; for(intI:rowina.keyset ()) { for(intJ:colinb.keyset ()) { for(intk = 0; K ) {Res[i][j]+ = Rowina.get (i) [K] *Colinb.get (j)

"Match link" click here~~"The main topic"n individuals, obtaining one to three degrees, each diploma has a size range, and the final diploma is awarded how many"Problem-solving ideas" are enumerated directly,Read the question, hurriedly wrote a code, found unexpectedly Miss registration time, System prompt can not submit code, really drunk ~ ~, later still have to register in advance:A problem, relatively simple:Code:#include Copyright NOTICE: This article for Bo Master original article, withou

"Topic link" click here~~"The main idea" gives you n boy,n a girl, then w represents the maximum capacity of the teapot, and then n Cups, each with a different capacity, requiring boy's cup of tea is twice times the girl, and boy and boy capacity, girl and girl capacity, like, ask how to pour tea, Maximizing the total amount of tea"Problem-solving ideas" This problem is very simple, the first read the question did not read, thought that each cup of tea is full, and then a think is not to take th

-Query output one line, containing either the word HAPPYOr the word UNHAPPY(Answers shocould be in the same order as the corresponding queries ). Sample Input sample input sample output ARRIVE 1 1ARRIVE 10 200BUY 5 900BUY 5 900BUY 5 1000 HAPPYUNHAPPYHAPPY There are two operations for a store: (1) arrive n C indicates that N items are purchased, each of which is c RMB. (2) Buy n t indicates that a buyer needs to buy n items, and a total of T yuan is spent. If the

complex. When the 3*3 is not finished, the remaining voids are also given priority to placing 2*2 as much as possible. When you place 1 3*3, you can also place 7 1*1 and 5 2*2 when you place 2 3*3, you can also place 6 1*1 and 3 2*2 when you place 3 3*3, and you can also place 5 1*1 and a 2*2 Because a 4*4,5*5,6*6 can only be placed in a box, so how many of these, it will take how many boxes to install. Then because the 3*3 can be placed in combination with the 1*1 and 2*2, it is also possibl

311-packets Time limit:3.000 seconds Http://uva.onlinejudge.org/index.php?option=com_onlinejudgeItemid=8category=113page=show_ problemproblem=247 A Factory produces products packed in square packets of the same height H and of the sizes,,,,,. These products are are always delivered to customers in the square parcels of the the same heighth as the products have and the Size. Because of the expenses it is the interest of the factory as as "the" the c

Nyist oj 311 full backpack (classic question of Dynamic Planning), nyistojFull backpack time limit: 3000 MS | memory limit: 65535 KB difficulty: 4 Description A full backpack defines N items and a backpack with a capacity of V. Each item has an unlimited number of items available. The volume of item I is c and the value is w. Solving which items are loaded into a backpack can make the total size of these items not exceed the total size of

; $ if(C 1) w[num]++; $ Elseb[num]++; - - for(inti =0; I ){ the intv =G[u][i]; - if(!Color[v]) {WuyiDFS (V,3-c,num); the}Else{ - if(Color[u] = =Color[v]) { WuF0 =true; - } About } $ } - } - - voidsolve () { A //3 sides + if(M = =0){ theprintf"%d%i64d\n",3, 1ll*n* (n1) * (n2)/6); - return ; $ } the for(inti =0; I ) g[i].clear (); thememset (Degree,0,sizeof(degree)); thef =true; the for(inti =0; I

(u->next[i]); u->sum+=u->next[i]->sum; } }}voidSeach (node *root) {node *p=root; while(1) {k-=p->flag;if(k0){return; }if(p->next[0]!=null) {if(p->next[0]->sum>=k) {str[cnt]=' A '; p=p->next[0]; cnt++;Continue; }Elsek-=p->next[0]->sum; } str[cnt]=' B '; p=p->next[1]; cnt++; }}intMain () {intI, J, Len, H; while(scanf('%s ', s)!=eof) {scanf("%d", k); len=strlen(s);memset(OK,0,sizeof(OK)); for(i=0; i1;if(i1s[i]==s[i+1]) ok[i][i+1]=1;if(i2s[i]==s[i+2]) ok[i][i+2]=1;if(i3

Topic: There are 1x1,2x2,3x3,4x4,5x5,6x6 of square blocks, ask at least a few 6x6 boxes can be put down.Analysis: Greedy heart. 1x1 is used to fill the remaining space, any space can be put down;First of all, each 6x6,5x5,4x4 to a single box, so the rest of the space to fill the 2x2,1x1 as the best;Then, the rest of the 3x3,2x2, first filled with the placement (4 3x3,9 2x2 Group assigned to the best);Finally, there are less than 4 3x3, 2x2 less than 9, up to two boxes, 1x1 last fill.(proving tha

*now= (!c?Cont1:cont2); Now[contcomp]++; for(intI=0; I) {flag=dfs (g[u][i],!c); } returnFlag; }}intMain () {///freopen ("/home/files/cppfiles/in", "R", stdin);Cin>>n>>m; for(intI=0; i){ intf,t; scanf ("%d%d",f,t); G[F].PB (t); G[T].PB (f); } if(m==0) {cout"3"1) * (n2)/6Endl; return 0; } BOOLflag=1; for(intI=1; iif(G[i].size () >1) flag=0; if(flag) {cout"2"2) *mEndl; return 0; } for(intI=1; i){ if(!Vis[i]) {Contcomp++; if(!dfs (I,0) ) {cout"0 1"Endl; re

Codeforces Round #311 (Div. 2) A, B, C, D, E A. Ilya and Diplomas Train of Thought: I have a question. I just need to enumerate it and discuss the situation as soon as possible. Code: #include #include #include #include #include #include #include #define inf 1000000#define clr(x, y) memset(x, y, sizeof x)using namespace std;int rmina,rmaxa;int rminb,rmaxb;int rminc,rmaxc;int n;int main(){ s

2 3*3, you can also place 6 1*1 and 3 2*2When you place 3 3 3*3, you can also place 5 1*1 and 1 2*2 Because a 4*4, 5*5, 6*6 can only be placed in a box, how many boxes are needed for installation. Then, because 3*3 can be combined with 1*1 and 2*2, you can determine the boxes required for 3*3. In the calculation, we can first determine the number of boxes after 3*3, 4*4, 5*5, and 6*6. Assume It is a T box. Then there will be gaps in the T box, and the 2*2 will be placed in the gaps in the T b

This paper introduces the application of the new ADO control in C++builder 5 environment, and gives a simple example. An overview of ADO ADO (Active Data Object) is a Microsoft Database model based on OLE DB. It implements a series of COM

UVA_311 This topic can directly simulate the packing process, and must first be large. ① Every 6x6 occupies a box. ② Each 5*5 is placed in a box, and 11 1*1 items can be installed in it. ③ Each 4*4 is placed in a box, and five 2*2 products can

13.6.1 statement Block The so-called statement block refers to multiple statements contained in curly brackets. The statement block is an overall execution body, similar to a separate statement. Code example: { A = 1; B = 2; Alert (A + B ); }   13.6.

Full backpack time limit: 3000 MS | memory limit: 65535 kb difficulty: 4 Description A full backpack defines n items and a backpack with a capacity of V. Each item has an unlimited number of items available. The volume of item I is C and