discrete earbuds

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

http://www.lightoj.com/volume_showproblem.php?problem=1085Test Instructions: the number of ascending subsequence sequences for a sequence.idea: to find the law can be found that a number as the number of the end of the type of the number of all smaller than its case +1. Use a tree-like array to find it. C++11 Auto is not available on Lightoj/.\/** @Date: 2016-12-01-21.58 * @Author: Lweleth ([emailprotected]) * @Link: https://github.com/* @Versi On: */#include Lightoj 1085-all Possible increasin

Segment tree + discrete IP address SEGMENT check SEGMENT TREE Problem: Give a series of IP segments, for example, [0.0.0.1-0.0.0.3], [123.234.232.21-123.245.21.1]... Now there is a new IP, find which IP segment it's in? Solution: First, we cocould map the ends of IP segments into some intervals, since the IP address cocould be represented by a unsigned int. This is calledDiscretization. Second, setup the segment tree. Every leaf node is [x, x + 1], it

Given p, k, a, where p, k is a prime number, x ^ k = a (mod p ). The legendary much simpler task .... If r is the original root of p (all prime numbers have the original root), then r ^ 1, r ^ 2... r ^ phi (p) constitutes the complete Residue System of the modulo p, so x = r ^ I, a = r ^ j can be set, then the equation is converted into r ^ (I * k) = r ^ j (mod p) then the theorem can obtain I * k = j (mod P-1) the r ^ j = a mod p can be obtained from the di