discrete flask

#include Main (){int i,n,j,k,a,b,x[100][100];while (scanf ("%d", n)!=eof){for (i=1;ifor (j=1;jscanf ("%d", x[i][j]);for (i=1;ifor (j=1;jfor (k=1;k{if (X[j][i])X[J][K]=X[J][K]+X[I][K];if (X[j][k])X[j][k]=1;}printf ("\ n");for (i=1;i{for (j=1;jprintf ("%d", x[i][j]);printf ("\ n");}}return 0;}Input40 1 0 00 0 0 10 0 0 01 0 1 0Output1 1 1 11 1 1 10 0 0 01 1 1 1Relationship closure with Warshall algorithm (discrete mathematics)

When there are too many other users 1 2 3 4 5 6 7 The new four intervals are [1, 3] [2, 4] [5, 7] [4, 6], and the overwrite relationship is not changed. The new number axis is 1 to 7, that is, the length of the original number axis is compressed from 9 to 6. Obviously, the line segment tree of [] is more space-saving than the line segment tree, the search speed is faster, but the results are consistent. Note that the same endpoint must be removed before sorting. This reduces the number of elem

Problem descriptionafter eating food from Chernobyl, DRD got a super power: He cocould clone himself right now! He used this power for several times. he found out that this power was not as perfect as he wanted. for example, some of the cloned objects were tall, while some were short; some of them were fat, and some were thin. More evidence showed that for two clones A and B, if a was no worse than B in all fields, then B cocould not keep ve. more specifically, DRD used a vector V to represent

Original question address: http://acm.timus.ru/problem.aspx? Space = 1 amp; num = 1010 The topic is defined in {1, 2, 3, 4 ..... the discrete function on n}, n After reading the questions, I naturally want to enumerate the connections between every two points. The time complexity is O (n square ). At the beginning, WA searched other people's code online and found that the AC code is the absolute value of the two-point difference. for example, for

In the non-grid method, ttttttttt draws discrete points. If there are many points, they cannot be represented by small balls. In my test, if the number of small balls exceeds one hundred, the speed is too slow to bear. The solution is to use the dot genie or glsl coloring language. # Include "vtkcamera. H"# Include "vtkrenderwindow. H"# Include "vtkrenderwindowindowinteractor. H"# Include "vtkinteractorstyleswitch. H"# Include "vtkpolydata. H"# In

the preceding step that the return value is when I * m + j> 0. So far, the above algorithms are not controversial, and the Code implemented by everyone is not much different. It can be seen that when C is a prime number, the problem of such discrete logarithm can become very easy to implement. #includeHunnu 10590 fzu 1352 PKU 2417 hit 1928 zoj 1898 /* a^x=b (mod c) c is prime*/#include#include#include#include#define LL long long#define nmax 46345typ

Bitmap Bitmap display. The following code displays an 8x8 chessboard: Glubyte WB [2] = {0x00, 0xff}; glubyte check [64*8]; for (INT I = 0; I 650) This. width = 650; "src =" http://s3.51cto.com/wyfs02/M02/48/10/wKioL1QFg__Bxp0fAACqcedUSYo765.jpg "Title =" qq20140902164614.png "alt =" wkiol1qfg1_bxp0faacqcedusyo765.jpg "/> Void glbitmap (glsizei width, glsizei height, glfloat x0, glfloat y0, glfloat XI, Glfloat Yi, glubyte * bits) // Draw a bitmap with width and height based on the array b

asked.Sample Input1051 2 Even3 4 odd5 6 even1 6 even7 OddSample Output3Title: There is a known length of 01 strings, given a number of conditions, [l,r] This interval in the number of 1 is odd or even, ask the first few are correct, no contradiction[L,r] The number of 1 can be expressed as sum[r]-sum[l-1], and the topic is only a parity, also determine the sum[r] and sum[l-1] parity is the same, to this step is simple, to change to the classic modelRange is a bit large, according to L-1,r Discr

is slightly different, for the Mij value, like the following three kinds of situations:(1) VI is not associated with EJ, mij = 0;(2) VI is the starting point of EI, mij = 1;(3) VI is the focus of EI, mij=-1;(4)Adjacency matrix of a directed graph: for AIJ, the number of edges that represent VI->VJ (direct connectivity)There is a very important theorem about the adjacency matrix of a graph:Set A is the adjacency matrix of directed graph D, the fixed point of D and v={v1,v2,v3...vn}, then the ele

Conditional probabilities:Probability of a occurring under the premise of B occurrenceP (a| b) =p (AB)/p (b)AB Mutual independence, P (AB) =p (A) *p (B)Bayesian formula:P (a| B) =p (b| A) *p (a)/P (B) [Shift]Full probability formula:Divides the sample space s into several disjoint parts (no repetition is not omitted)P (A) =p (a| B1) *p (B1) + ... "Probability of occurrence of each part * probability of occurrence of event a after occurrence"The formula is not important, so you can't restrict you

relative ER Ror of 10−6.Sample Input4 3 1 1 0.33 0.34 0.33 3 1 2 0.33 0.34 0.33 3 1 2 0.5 0.0 0.5 4 2 2 0.5 0.0 0.0 0.5Sample OutputCase #1:0.3300000Case #2:0.4781370Case #3:0.6250000Case #4:0.3164062 Each ball is independent of each other, and one can be asked.F[i] Indicates the probability of the total death of 1 balls and offspring after I daysF[i]=p[0]+p[1]*f[i-1]+p[2]*f[i-1]^2+ .....Boundary f[0]=0////main.cpp//uva11021////Created by Candy on 26/10/2016.//copyright©2016 Candy. All ri

width. Moreover, the candidates started placing their posters on wall segments already occupied by other posters. Everyone in Bytetown is curious whose posters would be visible (entirely or in part) on the last day before elections. Your task is to find the number of visible posters when all the posters was placed given the information about posters ' Size, their place and order of placement on the electoral wall. InputThe first line of input contains a number C giving the number of cases tha

Reflection:This problem reflects my analysis of this slag is not comprehensive, although 1A but after the discovery of a place to verify that they have neglected ...Code:#include intMain () {intp,k; intn=1e9+7; scanf ("%d%d",p,k); if(k==0) { Long LongTmp=p,r=1; P--; while(p) {if(p1) {R=r*tmp%N; } tmp=tmp*tmp%N; P>>=1; } printf ("%i64d\n", R); } Else if(k==1) { Long LongTmp=p,r=1; while(p) {if(p1) {R=r*tmp%N; } tmp=tmp*tmp%N; P>>=1; } printf ("%i64d\n", R); } Else

Discrete processing of coordinates

intinf=0x7fffffff;//infinitely LargeCharop[3];structinterval{intStart,endn; BOOL operatorConstinterval b)Const{ if(start!=B.start)returnstartB.start; Else returnendnB.endn; }};interval val[20001];voidWhiteintAintBintCNT) {Val[cnt].start=A; Val[cnt].endn=b; CNT++;}voidBlackintAintBintCNT) { inttmp=CNT; for(intI=0; i) { if(val[i].starta) {if(val[i].endn>=a) {if(val[i].endnb) {Val[i].endn=a-1; } Else{Val[tmp].start=b+1; Val[tmp].endn=Val[i].endn; TMP+

tree.The code is as follows.Private Boolean Judge (string[] relationstring, int[][] Matrix, int[] mark,int bef, int aft) {Find all pairings of the first element such as for (int i = 0; i if (matrix[aft][i] = = 1) {//Retrieve new elementif (i = = bef) {}//judge whether it is the edge that has been retrieved, judging is to jump directly on the lineelse if (bef! = i) {//Is the new edge retrievedif (mark[i] = = 1) {//with heavy edgesreturn false;}Mark[i] = 1;if (! Judge (relationstring, Matrix, mar

, transitive) closure is a relationship that satisfies the following conditions R ': (i) R ' is reflexive (symmetric, transitive); (ii) R '? R; (iii) for any reflexive (symmetric, transitive) relation on a R ", if R"? R, then there is R "? R '. The reflexive, symmetric, transitive closures of R are recorded as R (R), S (r), and T (r), respectively. Property 1 A two-tuple relationship on a set a closure operation of R can be compounded, for example: TS (r) =t (s (r)) represents the transitive clo

be visible (entirely or in part) on the last day before elections.Your task is to find the number of visible posters when all the posters was placed given the information about posters ' Si Ze, their place and order of placement on the electoral wall.InputThe first line of input contains a number C giving the number of cases that follow. The first line of data for a single case contains number 1 OutputFor each input data set print the number of visible posters after all the posters is placed.Th

Title: Given P,b,n, find the smallest L make b^l≡n (mod p) (p is prime)Naked bsgs ... Practice practiced hand--#include Bzoj 3239 Discrete Logging baby-step-giant-step

# Include # Include # Include Using namespace STD;Const int n = 20010;/* Is it PKU playing rogue or I opened too small? I originally followed 10010 by 10 * n, and the result is re, you can only pass n = 20010 !!!!! */ Int mark [10 * n];Struct tree{Int L, R; // The number of discrete pointsInt turn;} T [8 * n * 4]; Struct point{Int beg, end;Int turn;} Dia [N * 10]; Int node [10 * n];Int all; Void maketree (int c, int L, int R){T [C]. L = L;T [C