associate schedule

].sum[who] >= num)returnQuery (Lson, Num, who); if(A[lson].rsum[who]+a[rson].lsum[who] >=num)returnA[lson]. R-a[lson].rsum[who] +1; returnQuery (Rson, Num, who);}intMain () {intN, M, T, t =1, R, L, time; scanf ("%d", T); while(t--) {scanf ("%d%d", n, m); Build (1,1, N); printf ("Case %d:\n", t++); while(m--) { Chars[1100]; scanf ("%s", s); if(s[0] =='D') {scanf ("%d", Time ); L= Query (1, Time,0);///cock Silk is 0; if( ! L) printf ("Fly with yourself\n"); Else{

Week number Professional Learning Goals Professional Learning Time New Code Volume Blog post Volume Knowledge Summary 2nd week Contact data structure 2 hours 30 article 0 Unfathomable 3rd Week Data structure--linear table 4 hours 100 article 1 Read "I'm an IT little bird"Basic knowledge of learning data structures 4th Week Elements and attributes of HTML5 3 hours 20

Professional Learning goals Study time New Line of code Blog post Volume Humanities Learning Knowledge Summary fourth week Master learning 3 hours NBSP; 1 read "I am a It little bird", write the Reading book sequential storage structure for linear tables fifth week mastered more solid 3 hours NBSP; 1 read "The world is Digital", write the book review, learn

There is a total of n courses you have to take, labeled from 0 to n - 1 .Some courses May has prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a Pair[0,1]Given the total number of courses and a list of prerequisite pairs, are it possible for your to finish all courses?For example:2, [[1,0]]There is a total of 2 courses to take. To take course 1 should has finished course 0. So it is possible.2, [[1,0],[0,1]]There is a total of 2 courses to take.

then from 1 to J. (Note that the i>j. and not connected. ）For a random point I. There are two ways to connect, one is (a) to see, I is connected to the i-1, and the other is (b). I is not connected to i-1.Based on the two-tone journey. We know that the node n must be connected to N, so suppose we ask for the dp[n][n-1], just add it d[n-1][n] is the shortest two-tone closed route.Basis. Very easy to write equations:DP[I][J]=DP[I-1][J]+D[I][I-1];Dp[i][i-1]=min (Dp[i][i-1],dp[i-1][j]+d[j][i]);Here

TileImagery.prototype.processStateMachine is responsible for the loading of the image data; service satisfaction, to pay the bill: renderable = True/state = Quadtreetileloadstate.done.Through the above description, one of the Earth's rendering processes is mainly four links:L Grid DivisionL Terrain DataL Imagery DataL RenderingAnd these four links are staggered, although it seems a bit confusing, many functions, but do not know whether you notice, whether it is the rendering process or the upda

2016/9/12 MondaySend present to Teacher Weihu2016/9/13 TuesdayListen to the class as an assistant2016/9/14 Wednesday2016/9/15 Thursday2016/9/16 Friday2016/9/17 Saturday2016/9/18 SundayHold a meeting about Bnu-lsved 2.0###########################################Additional1.exercise/C + +2.exercise Java3.exercise python4.exercise Linux5.exercise algorithm6.exercise Mathematical Statistics###########################################Schedule of 2016-09-12~

Most important things to do1.joint PhD Preparations2.journal paper to write3.solid fundamental knowledge including 中文版4.body Fitness2016/10/31 Monday1.prepare ppt for this afternoon ' s group meeting2.join the Group meeting3.modify ppt for afternoon ' s group meeting and reflect myself4.make plans for this week###########################################Additional1.exercise/C + +2.exercise Java3.exercise python4.exercise Linux5.exercise Mathematical Statistics Machine learning algorithm########

Week number Professional Learning Goals Professional Learning Time New Code Volume Blog post Volume Learning in the Humanities Knowledge and Skills Summary 4 HTML:4 hoursData:　　　　2 hours 69 1 Articles:Summary of the week four Xin Zou Teacher's modern software engineeringThe software engineering education of learning"The World is digital" HTML:Creation of dynamic tablesData:Static storage 5 Html:Make a lit

Psp Category C Content C Start Time S End time t Interrupt Time I NET time t Activities Stand-up meeting this week —— —— —— 100 This week's documentation —— —— —— 120 This week's code related —— —— —— 320 Other project matters —— —— —— 150 Line of code 1350+210=1560

: represents the day of the week, which can be any integer from 0 to 7, where 0 or 7 represents Sunday. Command: The command to execute can be either a system command or a script file that you write yourself. In each of the above fields, you can also use *,-/special characters, which are described below:650) this.width=650; "src=" Http://s5.51cto.com/wyfs02/M01/80/0E/wKioL1c1x2nCvW3BAAAqgA7k9ec790.png "title=" capture. PNG "alt=" Wkiol1c1x2ncvw3baaaqga7k9ec790.png "/>3. Example implem

you need to know as soon as possible that this is not clear! and step by step to do it never clear to clear, it depends on your comprehensive ability, and play the real-time strategy game when the full picture Allbright which is the comprehensive experience of strategic and tactical integration strength ...Iii. SummaryIn fact, that is the exchange, communication, collaboration, such issues are not in fact a separate software development, no matter what kind of events are more than one person ma

Moonphases Study time Newly written lines of code Blog Volume (article) Learn the point of knowledge 1 10 1 Software 2 3 4

Study time New Line of code Blog Volume Knowledge Summary 1th Week 2nd week 3rd Week 4th Week 5th week 6th Week 7th Week 8th Week 9th Week

First, homework (exercise) content:1. Summarize the use of the commands involved in this course and show the relevant examples;2. Summarize basic regular expressions and extended regular expressions3. Display the lines in the/etc/passwd file that end with bash4. Display the two-digit or three-digit number in the/etc/passwd file5. Display lines with ' LISTEN ' followed by 0, one or more whitespace characters in the ' Netstat-tan ' command result6. Add user bash, Testbash, basher, and Nologin user

first behavior of each group of data is the integer n (1OutputThe output can get the maximum boost effect.One row for each set of test data output.Sample Input232 5 54 6 46 7 1031 2 32 4 103 10 2Sample Output1510SourceEast China Normal University 2009 school raceProblem solving: Direct DP OK, dp[j] means that in J time, can get the maximum promotion effect is how much, then similar to 01 backpack, a blind1#include 2 using namespacestd;3 Const intMAXN = +;4 intDP[MAXN];5 structnode{6 ints,t,

have first course, that is, a course may have to finish the B course after the election, ask for a legal elective these classes of the order, no legal solution to return empty.Problem Solving Ideas:Topology sequencing, can be done by BFS, first of all, each course has a degree, into the degree of 0 as the last elective, and then it depends on the degree of the course to reduce one, and then handed over to all classes are selected. Note that this problem data is relatively large, can not use two

impatient air ... Learning programming language requires a good foundation of English as a guarantee. My own English is very poor, I suffer from it.Finally, the study schedule is listed below for reference only:1, Basic Learning: installation configuration and other Basics : 3 months.2, Read and fully understand the code of others , familiar with object-oriented 12 months.3, hold other technologies : include: Javascript,xml,ajax,mysql,http protocol.

[flag][v].ma = Tree[flag][v].ma_r = (tree[flag][v].r-tree[flag][v].l + 1) *tag;return;}Push_down (v);int mid = (tree[flag][v].l + tree[flag][v].r) >> 1;if (a if (b > Mid) update (A, B, right, flag, tag);Push_up (v);}int main (){Freopen ("In.txt", "R", stdin);int T;scanf ("%d", t);int cas = 0;while (t--){int N, Q;scanf ("%d%d", n, q);Build (1, N, 1);Char ch[10];int L, R, QT;printf ("Case%d:\n", ++cas);while (q--){scanf ("%s", ch);if (ch[0] = = ' D '){scanf ("%d", AMP;QT);if (Tree[1][1].ma {Puts (

a predecessor course (a course with a 0 degree) into the queue + for(inti =0; I ) A if(grap[i].size () = =0) at Queue.push (i); - - while(!queue.empty ()) - { - //Take out the team's first class - intCur =Queue.front (); in Queue.pop (); -Traverse the current course to //traverse the post-course of the current course (cur), removing the current course from the set of pre-courses (Grap[*it]) in the post-curriculum (*it) +