at t monroe la

Topic PortalTest instructions: The small island of the convex polygon in the sea, ask the island's point to the sea farthest distance.Analysis: Training Guide P279, two-point answer, and then the entire polygon to internal contraction, if the half-plane intersection non-empty, then these points constitute a half-plane, there are satisfied points./************************************************* author:running_time* Created time:2015/11/10 Tuesday 14:16:17* Fi Le Name:LA_3890.cpp ***************

output for three test cases.Sample Input3 15 4 10 2 7 7 5 12 2 7 12 4 9 13 6 3 2 1 3 4 0 2 5 4 6 10 4 8 2 5 4 2 5 6 5 8 2 5 8Sample OutputYES NOYESThe main idea: to give you a lot of targets, each target is an interval and has its own height, ask you can in a given interval to find a straight line to all these targets strung together.Analysis: Look at the online solution to know the practice, two points archery position, and then maintain a section of elevation, if a target in the current inter

Interval update + statistical update length Just a little bit of attention, T. #include #defineLson L, M, rt#defineRson m+1, R, rtusing namespacestd;Const intMAXN =100000+131;intma[maxn2], mi[maxn2], lazy[maxn2];intCnt;voidPushup (intRT) {Ma[rt]= Max (ma[rt1], ma[rt1|1]); MI[RT]= Min (mi[rt1], mi[rt1|1]);}voidPushdown (intRT) { if(LAZY[RT]! =-1) {Lazy[rt1] = lazy[rt1|1] =Lazy[rt]; Ma[rt1] = ma[rt1|1] =Lazy[rt]; Mi[rt1] = mi[rt1|1] =Lazy[rt]; LAZY[RT]= -1; }}voidBuild (intLintRintRT) {Ma[rt]=

first instruction that gives the coordinates of the starting position. You may assume there is no more than instructions in each test case, and all the integer coordinates is in the range (-300, 300). The input is terminated if N is 0.OutputFor each test case there'll be one output line in the formatCase X:there is w pieces.,where x is the serial number starting from 1.Note:the figures below illustrate the and the sample input cases. Sample Input50 0 0 1 1 1 1 0 0 071 1 1 5 2 1 2 5 5 1 3

Test instructions: Given a sequence of DNA of length n of M, a shortest sequence of DNA is obtained, which minimizes the total hamming distance.Hamming the distance is equal to the number of different positions of characters.Analysis: See this problem, my first feeling is to calculate the time complexity, good small, nothing, completely can violence, as long as the same position on each string,The choice appears most, if has the same choice ASIIC code small (because requires the dictionary order

-Rujia's white paper describes the topic./*problem:status:by wf,*/#include "algorithm" #include "iostream" #include "CString" #include "Cstdio" #include "string" #include "stack" #include "cmath" #include "queue" #include "set" #include "map" #define Lson l, M, RT 　　LA 3942 Remember The Word (trie tree)

*q =NewLine[n]; Aboutq[first=last=0] = l[0]; the for(intI=1; I) the { the while(First1])) last--; + while(First; -Q[++last] =L[i]; the if(Fabs (Cross (Q[LAST].V, q[last-1].V)) EPS)Bayi { thelast--; the if(Onleft (Q[last], l[i]. P)) q[last]=L[i]; - } - if(first1] = Getintersection (q[last-1], q[last]); the } the while(First 1])) last--; the //Delete a useless plane the if(last-first1)return 0; -P[last] =getin

#include #include#includeusing namespacestd;Const intMAXN =4000* -+Ten;Const intSigma = -;Const intMaxnn =300010;Const intMoD =20071027;Charwords[Maxnn];intdp[Maxnn];structtrie{intCh[maxn][sigma]; intVALUE[MAXN]; intsz; Trie () {sz=1; memset (ch[0] ,0,sizeof(ch[0]) ); } voidInit () {sz =1; memset (ch[0] ,0,sizeof(ch[0]) ); } intDixCharc) {returnC'a' ; } voidInsert (Char*s) {intLen =strlen (s); intU =0; intI, cur; for(i =0; i ) {cur=dix (s[i]); if( !Ch[u][cur]) {memset (Ch[sz],0,sizeof(Ch[

not nuclear. That is, the angle sequence can not have two o adjacent, and the tail can not be at the same time O.Then we determine the number of R and O. The internal angles of the N-edged shape are (N-2) *180, and R represents the inner angle of 90,o for 270, so it is easy to calculate the number of R and O, respectively (N + 4)/2 and (n-4)/2. The problem is then converted to how many methods of placing (n-4)/2 O and any two nonadjacent are placed in n positions.This is a very simple combinato

#include 　　[And look up the set] LA 3644 x-plosives

Title Link: Https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudgeItemid=8page=show_problem problem=4846The main problem: the street equidistant distribution of n stores, numbered 1~n, the distance between adjacent stores is 1, someone from the initial position (0 points) to buy some items in each store, and shopping to meet some such as first to store a, then to store b restrictions, and finally reached the exit (N+1 point). Ask this person what the shortest way to go for a shoppin

characters long. There'll be no, identical words and all letters in the words would be lowercase.There is a blank line between consecutive test cases.You should proceed to the end of file.OutputFor each test case, the output of the number, as described above, from the task description modulo 20071027.Sample InputABCD 4 a B cd ABSample OutputCase 1:2Test instructions: Given some words, and a long string, ask this long string split into existing words, can split into several waysIdeas:Dp[i]=sum (

+ (B-A)/2; One return(f (a) +4*f (c) +f (b)) * (B-A)/6; A } - DoubleAsrDoubleADoubleBDoubleEpsDoubleA) - { the Doublec=a+ (B-A)/2; - DoubleL=simpson (a,c), r=Simpson (c,b); - if(Fabs (l+r-a) the*eps)returnl+r+ (L+R-A)/15.0; - Else returnASR (a,c,eps/2, L) +asr (c,b,eps/2, R); + } - intT; + intd,h,b,l; A intMain () at { - //freopen ("1.in", "R", stdin); -scanf"%d",T); - for(intKase=1; kaseKase) - { -scanf"%d%d%d%d",d,h,b,L); in intN= (b+d-1)/D; - Doub

greedy to find the best strategy. There is an explanation on the white. But here are a few places to think about.　　There are a few places to understand, the brain is not enough, first remember, another day to do it again#include #includeusing namespacestd;Const intMAXN =100005;intN,A[MAXN],LEFT[MAXN],RIGHT[MAXN];BOOLOkintp) { intx = a[1],y = p-a[1]; left[1] = a[1];right[1] =0; for(inti =2; ii) {if(i%2==0) {Left[i]= Min (a[i],x-left[i-1]); Right[i]= a[i]-Left[i]; } Else{Right[i]= Min

Topic PortalTest instructions: South of Training (P216)Analysis: To find the most occurrences of the string, then each string map id,cnt record the maximum number of times can be.#include 　　AC automaton LA 4670 dominating Patterns

Test instructions: Given a binary tree, and numbering and stipulating each one, now give you a value that asks you what the first few.Analysis: This problem, I think for a long time to come out, this is really the data structure is too poor, not solid enough, this problem, should be pushed from the bottom up, if the numerator is greater than the denominator, then this number is odd,To add 1, if it is less than, do not add. Just push on to the first one.The code is as follows:#pragma COMMENT (lin

of 4.InputThe input contains several test cases. The first line of the input contains a positive integerZ, denoting the number of the test cases. ThenZ -Test cases follow, each conforming to the format described below.The input is one line containing a word's length at most 3.consisting of (large or small) letters of the E Nglish alphabet.OutputFor each test case, the your program have to write a output conforming to the format described below.You should output one integer K being the power of

Some words are known, select some of the words to form the purpose string, ask how many methods. In fact, at the beginning of this problem, it is natural to think of the classic coin problem in dynamic planning: For example, ask 1 yuan, 2 yuan, 5 yuan, total how many ways can constitute 20 yuan? This is just a change of the coin to a word. However, if it is really just like the coin problem, each word is polled, obviously too slow, up to 300000*4000*100 times.If you use the trie number, you can

Five common la s of Swing and wing in java 1. Border layout (BorderLayout) 2. FlowLayout) 3. GridLayout) 4. BoxLaYout) 5. Empty layout (null) There are two other layout S: GridBagLayout (grid package layout) and CardLayout (card layout) Note: The default layout of JFrame and JDialog is BorderLayout, while that of JPanel and Applet is FlowLayout. Sample Code for border layout: Import java. awt. borderLayout; import javax. swing. JButton; import javax.

Test instructions: Given n individuals, and their number is compared, starting from behind, if not long enough, add middle then than, until not, if the same than the name.Analysis: The problem of water, it is needless to say.The code is as follows:#pragma COMMENT (linker, "/stack:1024000000,1024000000") #include 　　LA uvalive 7370 Classy (sort, compare)