nex machina

1. Getting Started with Java　　Java: A high-level object-oriented programming language introduced by Sun (Stanford University Network, Stanford University Inc.) in 1995.　　SUN:2009 was acquired by Oracle (Oracle Corporation).　　James Goslin: The father of Java.　　Java History:2004 j2se1.5 release, is a milestone in Java development. and renamed to Java SE 5.0;2005 Sun Company Public Java SE 6.0,java various versions are renamed;J2ee→java EE;J2se→java SE;J2me→java ME;Acquired by Oracle (Oracle Corpor

http://poj.org/problem?id=3041There are K asteroids in the N*n grid, where the asteroid I is (RI,CI), and now there is a powerful weapon that can blast a whole row or a whole array of asteroids to ashes with a luminous speed, and to use this weapon to destroy all asteroids requires at least a few luminous beams.Mainly the composition, will each line into a point, constitute a set of 1, each column also as a point, the composition of the set 2, each obstacle position coordinates will set 1 and se

the longest route to the hotel. Note that there must be no forks on the road and the road is backward. Direct dfs + pruning on the line, the graph is saved with set... Look at others using a one-dimensional array. Code: /** Author: illuz * Blog: http://blog.csdn.net/hcbbt* File: b.cpp* Create Date: 2013-10-02 00:17:29* Descripton: b */#include #include #include #include #include using namespace std;#define rep(i, n) for (int

:#include #include #include #include using namespace STD;inthead[60005],val[160005];inttot,to[160005],nex[160005];intNintAA,BB,CC; Queueint>Qintd[60005];intvis[60005];voidSPFA (intx) {memset(d,-1,sizeof(d)); d[x]=0; vis[x]=1; Q.push (x); while(!q.empty ()) {intNow=q.front (); Q.pop (); vis[now]=0; for(inti=head[now];i!=-1; I=nex[i]) {intV=to[i];if(d[v]==-1|| D[v]if(!vis[v]) {vis[v]=1;

characters in "AGCT", which is the DNA segments causing inherited disease.The last line of the that test case is a non-empty string of length not greater than-containing only characters in "AGCT", Which is the DNA to be repaired.The last test was followed by a line containing one zeros.Outputfor each test case, print a line containing the "test Case number" (beginning with 1) followed by theNumber of characters which need to be changed. If it ' s impossible to repair the given DNA, print-1.Samp

DescriptionThe farmer John's farm suffered an earthquake. Some of the cattle sheds were damaged, but fortunately, all the road stalls between the cattle can be used. FJ's Farm has P (1 Input* Line 1th: Three spaces separate number: P, C, and N* 2nd. C+1 lines: Two spaces per line: a_i and B_i * c+2..c+n+1 line: One number per line: Report_jOutput* Line 1th: Number of cows that cannot return to the farm at least (including damaged barns).Sample Input4 3 11 22 33 43Sample Output3HINTThe Barn 2 was

: +End_index = Self.total_count + 1 AStart_index = self.total_count-self.pager_num + 1 the + ifSelf.current_page = = 1: -Prev ='' $ Else: $Prev =''% (Base_url, self.current_page-1,) - page_list.append (prev) - the forIinchRange (int (start_index), int (end_index)): - ifi = =Self.current_page:Wuyitemp =''%(Base_url, I, i) the Else: -temp =''%(Base_url, I, i) Wu page_list.append (temp) - About ifSelf.current_page = =Self.total_count: $

Test instructions: To a city to get money, to leave to spend money. There is cash at the beginning. The city is circular, ask from which start, can on the way any time money >=0;A start pointer I, a trailing pointer J. Indicates an interval. If the conditional ++j is met and the proceeds are added to the sum (the earnings may be negative). Non-conforming is ++i, and the return from sum proceeds until the sum>=0; interval length is n, the position of I is the result.Proof of correctness: Suppose

2#include 3#include Set>4#include 5#include 6#include 7#include 8#include 9#include string>Ten#include One#include A using namespacestd; - #defineINF 0x3f3f3f3f - the Charp[100010],s[100010]; - intnex[100010]; - intLens; - + void Get(Char*p) - { + intplen=strlen (p); Anex[0]=-1; at intk=-1, j=0; - while(J Plen) { - if(k==-1|| P[J] = =P[k]) { -++J; -++K; - if(P[j]! =P[k]) innex[j]=K; - Else tonex[j]=Nex

];isum; - } -l[sum]=r[sum-1]+1; r[sum]=en; - for(intI=l[sum];isum; in } - voidInsert (Const intAMP;X) {++b[x]; + +sumv[num[x]];} to voidDelete (Const intAMP;X) {--b[x];--sumv[num[x]];} + intNext (Const intx) - { the for(inti=x;iif(B[i])returni; * for(inti=num[x]+1; iif(Sumv[i]) $ for(intj=l[i];j)Panax Notoginseng if(B[j])returnJ; - return-1; the } + intMain () A { the R (n); R (m); + for(intI=1; ii) a[i]. Read (); - for(inti=n+1; ii) $ { $

integer for each test case:the lowest intersection number @ which a new fire station SH Ould is built so as to minimize the maximum distance from any intersection to the nearest fire station. Sample input 1 621 2 102 3 103 4 104 5 105 6 106 1 10 Sample output 5 Source University of Waterloo Local Contest 199 Uploaded by Zhang Yunzun Helples

Iteration:Code:/*** Definition for singly-linked list.* Public class ListNode {* int val;* ListNode Next;* ListNode (int x) {val = x;}* }*/public class Solution {Public ListNode reverselist (ListNode head) {if (head = = NULL) return null;ListNode tail = head.next;ListNode cur = head;while (tail!= null) {ListNode NEX = Tail.next;Tail.next = head;Cur.next = NEX;head = tail;tail =

1OutputNOInput4 7.. x.xx.. Xx.. X.X... X..X......2 21 6OutputYESTitle, every step, go through that place into a broken ice, if you go again will fall, you can not beat in situ to make the ice change state, and your final goal is not just to find the destination, and to make it into X and then re-enterOriginally I wanted to think about the situation, such as the destination is X or. , and then judge the condition is changed, and then count, mark is not a problem, and consider from the origin to

Struct node {int data; node * Next;}; // create a linked list and enter a number as the input. If the input is 0, the linked list ends (0 is excluded) node * creat () {node * head, * P, * s; int X, cycle = 1; head = (node *) malloc (sizeof (node); P = head; while (cycle) {cout // Calculate the length of the linked list int computlength (node * head) {node * P; int n = 0; If (Head = NULL) {return 0;} p = head; while (P! = NULL) {n ++; P = p-> next;} return N ;}// Print the data void printlist (no

Http://poj.org/problem? Id = 3256 Store the paths of N farms in an adjacent table, record the initial number of cattle in each farm, and query the number and number of cattle in each farm along the path DFS. Code:# Include # Include Using namespace STD; Int num [1001]; Int sum [1001]; Int head [1001]; Bool vis [1001]; Int X; Struct past { Int V, NEX; } Edge [2, 10001]; Void addedge (int u, int v ){ Edge [X]. V = V; Edge [X].

) scanf ("% s",)# Define SLD (a) scanf ("% LLD", )# Define PF (a) printf ("% d \ n",)# Define PPF (a, B) printf ("% d \ n", a, B)# Define SZ (ARR) (INT) A. Size ()# Define swap (A, B) A = a xor B; B = a xor B; A = a xor B;# Define read freopen ("in.txt", "r", stdin)# Define write freopen ("out.txt", "W", stdout)# Define max 1 # Define ESP 1e-5# Define lson L, M, RT # Define rson m + 1, R, RT Template Template Template Template Template Const int maxn = 55555;Int Loy [maxn Int L [maxn], R [maxn];

HDU 4411 Arrest fee stream Question link: Click the open link Question: Given n + 1 vertex ([0, n]) an undirected graph of m edge. The starting point is 0, and k people are initially at the starting point, Traversing the graph enables at least one person to walk each vertex and the I-1 must have been traversed when traversing the I point. To minimize the path and number of k people, k people should return to the starting point. Ideas: Cost flow, because for a person, the sequence of the person's

# Include # Include # Include # Define eps 1e-9 # define pi acos (-1.0) using namespace std; # define ll int # define inf 0x3f3f3f # define Inf 0x3FFFFFFFFFFFFFFFLL # define N 105 # define M 1005 struct Edge {ll from, to, cap, cost, nex; Edge () {} Edge (ll from, ll to, ll cap, ll cost, ll next): from (from), to (to), cap (cap), cost (cost ),

than zero is the solution. # Include # Include # Include # Include # Include # Include # Include # Include # Include # Include # Include # Include # Include # Include # Define sqr (x) * (x )) # define LL long # define itn int # define INF 0x3f3f3f3f # define PI

(that is, the end element of the table points to the header node ), two-way linked list (elements can be quickly located to find their precursor and successor ). Common Operations on linked lists 1. Reverse of a single-chain table. // CCode: single-chain table implementation: Structure pointer typedef struct node {elemtpye data; struct node * Next;} node; typedef struct node * linklist; linklist * reveral (linklist * ls) {// The linked list is empty or if (LS = NULL | LS-> next = NULL) {return