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To give you a grid with a height of n and a width of M, to figure out how many rectangles there are in this grid.Here we can see this:For rows: If it is M-unit length, the rectangle with length 1 has m, and the rectangle of length 2 has m-1 ... The length of M is only 1, so the total rectangle is 1+2+...+m= (1+m) *M/2For columns: peers can.So, the total number of rectangles = row * column. The general formula N (m,n) = (1+m) * (1+n) *M*N/4 is obtained.#include usingnamespace std; int t,n,m; int

the starting point can arrive, to relax this edge even two points, do not know why need n-1 times can relax ...FloydFeel this more violent, the direct triple loop to traverse the number of n, K cycle n times, is to see whether K this point can shorten the weight, can shortenUse the first estimate will not time out. Initialization is the addition of the read-in Edge, the d[i][i]=0SpfaAn algorithm that is seen on the net is generally slower than Dijkstra.My favorite, feel like Bellman-ford optimi

Test Instructions Summary: The last small a, with your help, succeeded in separating the first part of the sage array-the triangle-and now he is ready to draw the second part of the law-the MO-Test rectangle.And he came across a problem, and he didn't know how many different rectangles there were.Uphold the help of the end, send the Buddha to the west of the Basic Law, you need to tell him the number of different rectangles.Problem Analysis: If the rectangle length is 1, the width is m, then the

^3 = a^2 + b^2+ c^2 = A+b+c = 1$, then $abc =? $Answer:by $ $a ^3 + b^3 + c^3-3abc = (A + B + C) (a^2 + b^2 + c^2-ab-bc-ca) $$ and $$ (A + B + c) ^2-(a^2 + b^2 + c^2) = 0$$ $$\ RightArrow 1-3ABC = 1-(ab + BC + CA) = 1\rightarrow ABC = 0.$$5. If $2a = 6b = 3c$, and $ab + BC + CA = 99$, then $2a^2 + 12b^2 + 9c^2 =? $ Answer: $$\begin{cases}a = 3b\\ c = 2b \end{cases} \rightarrow 3b^2 + 2b^2 + 6b^2 = \rightarrow b^2 = 9.$ $ so $$2a^2 + 12b^2 + 9c^2 = 18b^2 + 12b^2 + 36b^2 = 594.$$6. Set $x-y = 1+m$

http://www.lydsy.com/JudgeOnline/problem.php?id=4104I'm too lazy to write on the internet#include Good abuse of the heart of a problem, looked at the night Qwq"Bzoj 4104" "Thu Summer Camp 2015" decryption operations

output7TipsData rangen"Analysis" double pointer#include #include#include#include#include#includestring>#include#include#include#include#defineINF 2e9#defineMet (b) memset (a,b,sizeof a)typedefLong Longll;using namespacestd;Const intN = 1e6+5;Const intM = 4e5+5;intn,m;intL=0, r=0;intcnt[n],sum=0, ans=inf;intA[n];intMain () {scanf ("%d%d",n,m); for(intI=0; i) {scanf ("%d",A[i]); } for(intI=0; i) {Cnt[a[i]]++; R=i; if(cnt[a[i]]==1) sum++; if(sum>=m) { for(intj=l;j) {Cnt[a[j]]--; i

respectively indicate the time the ship arrived at the seaport and the number of passengers on board, and then ki an integer x (i,j) indicates the country of the passenger on board 7.Ensure that the input ti is incremented, the unit is the second, indicating that the first time from the beginning of the work of K, this ship in ti seconds to reach the seaport.Guarantee.Which represents all the Ki's and.Output format:Output n lines, line I outputs an integer representing the statistics of the arr

() {returnName } Public int Getage() {returnAge } Public Boolean equals(Object obj) {if(! (objinstanceofperson))return false; Person P = (person) obj; System.out.println ( This. name+"......"+p.name);return This. Name.equals (P.name) This. age==p.age; }}Class ArrayListTest2 { Public Static void SOP(Object obj) {System. out. println (obj); } Public Static void Main(string[] args) {ArrayList Al =NewArrayList ();//al.add (Object obj); Object obj = new Person ("Zhangsan", +);Al.add (N

/wKiom1csOgXAR0UOAAEaxRQIgIM969.jpg "title=" 3.jpg " alt= "Wkiom1csogxar0uoaaeaxrqigim969.jpg"/>The first class meeting simple but contains profound meaning, not only help everyone to eliminate a day of learning fatigue, but also let the small partners more understanding of each other, enhance unity and cohesion, and determine the classes leader to facilitate the future class management and activities planning arrangements. What a perfectly formed! Here we have to give the best of the beauty of

#include #include #include #include #include #include #include #include #include #include #include using namespace Std;typedef long long ll;typedef unsigned long long ull; #define MM (A, B) memset (A,b,sizeof (a)); const Double EPS = 1e-10;const int inf =0x7f7f7f7f;const double pi=acos ( -1); int p2,q2,p1,ep,q1;const int inf = 0x3f3f3f3f;const int MAXN = 1100;int L

intI=0; Atime++; - while(SCANF ("%d%d", AMP;A[I].I,AMP;A[I].J) (a[i].i! =-1|| A[I].J! =-1)) - { theNumber =i; - if(A[i].i | |A[I].J) -i++; - Else + return 1; - } + return 0; A } at voidtree () - { - intnow,next,q; - node y; - if(A[NUMBER].I = =0 A[NUMBER].J = =0 Number = =0) - { inprintf ("Case %d is a tree.\n", time); - return; to } + for(intI=0; i) - { the for(intk=0; k) * { $ if(A[I].J = =

The silly question of finding the number of reverse order.Use a merge sort or tree array or line segment treeMerge sort code:# include　　2016 Summer Camp Training 2 H q frosh Week

(a>0) - { -s + =Flower[a]; +A-= A -A; - } + returns; A } at intMain () - { - intt,a,b; - Chars[ A]; -scanf ("%d",t); - for(intj=1; j) in { -scanf"%d",n); tomemset (Flower,0,sizeof(flower)); + for(intI=1; i) - { thescanf ("%d",b); * Add (i,b); $ }Panax Notoginsengprintf"Case %d:\n", j); - while(SCANF ("%s", s) s[0] !='E') the { +scanf ("%d%d",a,b); A if(s[0] =='A') the Add (A, b); + if(s[0] =='S')

Silly greedy problem.First, according to the order of the points from the big to the small, the same score according to the cutoff date from small to large order. Then, in order, start looking for the time that is not occupied from the due date. If you cannot find it, add it to the penalty points.# include 　　2016 Summer Camp Training 2 I question Doing homework again

[ -];voidQueue_push (intx) {queue[++top] =x;}intQueue_pop () {returnqueue[under++];}intMain () {inti,j; for(i =0; I 5; i++) { for(j =0; J 5; J + +) {cin>>A[i][j]; }} memset (visit,0,sizeof(visit)); Queue_push (0); visit[0][0] =1; while(undertop) {v=Queue_pop (); if(v = = -) Break; intx = v/5, y = v%5; for(inti =0; I 4; i++) { intPX = x+dir[i][0],py = y +dir[i][1]; if(visit[px*5+py][0] ==0a[px][py] = =0px>=0px5py>=0py5) {Queue_push (px*5+py); satisfies the co

-bodied state that can't be given before the work. At my age, participating in training institutions to study the students have been very few, if not to find their professional interests, not to find the huaqing vision, I may not give up the work of the ease of the previous, but also did not have the courage to give up after-tax 10k salary. But I now believe that the salary I can get in the future can definitely surpass this!In the future, the greatest charm is that there are countless possibili

output is no. In addition there is an empty line behind each corresponding block. The output file does not exist in response to the last what data does not have. Output51 2 3) 4 55 4 1) 2 3066 5 4 3 2 100OutputYesNoYesProblem Analysis: Transit C, carriages conform to the principle of LIFO, you can use the stack to solve the problem, it is worth noting that to enter the possible outbound order, rather than the stack order. 1#include 2#include 3#include 4 using namespacestd;5 intn,t[1005];6 intMa

programmer gathers together to develop the program without guidelines, it appears as phpBB, Phpnuke, and a lot of rough. PHP3 packages. But can you simply blame the PHP developer? No! Other Web language giants, ASPs and Perl, are also notorious for developing a style of stew. So that's why PHP gets such a stigma. is because of its historical spread. Most PHP developers who have turned to Python,ruby and Java have not gone back to rethinking the language as the MVC concept arose. In addition, th

A Cold!!! ： （Morning listening to picks introduction to the polynomial (all content is not tested) only understand the O (n1.585) polynomial multiplication algorithm: (The bomb screen is installed! It's amazing (all the computers are in the brush screen): (Good lunch:) slept for half an hour afternoon nap actually fell asleep: |Listen to Ju Yifu in the afternoon to speak formal language and automata (all content is not tested). Beginning to feel able to understand, quite interested, after listen

{ -Pcur = *Root; A while(1) + { the if(Pcur->val >elem) - { $ if(pcur->lchild==NULL) the { thePcur->lchild =pnew; the Break; the } - Else { inPcur = pcur->Lchild; the Continue; the } About } the if(Pcur->val elem) the { the if(pcur->rchild==NULL) + { -Pcur->rchild =pne