algo vpn

The problem description S01 string is initially " 0 " Transform 0 Change to 01 input format 1 integers (0~ +) output format N-transform S01 String Sample input 3 Sample Output 101 data size and conventions 0~Remember:The information given by the topic is misleading, so people can't see the law (1 should be "1")The specific requirements are obtained by looking at the results of the test data (Lll￢ω￢)That is: the string of n = (n-1 string) + (N-2 string)AC Code:1#include 2 #defineMAX 100103 4 /*de

The problem description enters a sequence of 10 integers that requires ascending sorting and removes duplicate elements. Enter a format of 10 integers. Output format multiple lines of output, one element per line. Sample input 2233115555 Sample Output 1235Problem Solving Ideas:If the input number exists in the array, it is excluded, otherwise the storageSort from small to large, and outputAC Code:1#include 2 #defineMAX 103 4 intMainvoid)5 {6 inti =0, j =0, k =0;7 intarr[max+1];8 fo

Algorithm training Delete Array 0 element time limit: 1.0s memory limit: 512.0MB Reads n integers from the keyboard into the array, writes the function compactintegers, deletes all elements with a value of 0 in the array, and then moves the elements to the first end of the array. Note that the Compactintegers function needs to accept the array and its number of elements as arguments, and the function return value should be the number of new elements of the array after the delete operation execut

); + A System.out.println (maxsum); at } - /** - * Maximum and - * @paramRow number of rows - * @paramarr Array - * @returnthe path of the maximum and in */ - Private Static intMaxsum (intRowint[] arr) { to for(inti = row-2; I >= 0; i--){ + for(intj = 0; J ){ -ARR[I][J] + = Maxnum (arr[i+1][j), arr[i+1][j+1]); the } * } $ returnArr[0][0];Panax Notoginseng } - the /** + * Find the maximum numbe

+ intm = y-2 > 0? y-2:0;//the starting position of the Emperor's column range ax = x > n-1? N-1: x;//the end position of the Emperor's line range aty = y > n-1? N-1: y;//the end position of the Emperor column range - -Column_num =New int[n]; -Palace =New int[n][n]; - - for(inti = k; I //Emperor Range-1 in for(intj = m; J ){ -palace[i][j] = 1; to } + } - thePlace (0); * $System.out.println (count);//number of o

((src.charat (k) = = ' + ' | | | Src.charat (k) = = '-') (temp.length ()! = 0)) the break; the the temp.append (src.charat (k));94 Src.deletecharat (k); the } the they =integer.parseint (temp.tostring ());98Temp.delete (0, Temp.length ()); about - Switch(OP) {101 case+:102sum = x +y;103Src.deletecharat (k-1);//Delete operator104Src.insert (k-1, sum + "");//the result is stored in the operator position of SRC (the numbe

representation of the number of bits. added: Decimal to 16 binary: integer.tohexstring (int i)decimal Turn octal: integer.tooctalstring (int i)decimal Turn binary: integer.tobinarystring (int i)hexadecimal to decimal: integer.valueof ("FFFF", +). ToString ()Octal decimal: integer.valueof ("123", 8). ToString ()Binary to decimal: integer.valueof ("1001", 2). ToString ()Example code:1 ImportJava.util.Scanner;2 3 Public classMain {4 Public Static voidMain (string[] args) {5Scanner sc =NewSca

Algorithm training 4-3 Daffodils number time limit: 1.0s memory Limit: 256.0MB problem description Print all daffodils between 100 and 999. The so-called Narcissus number refers to the number of cubes that satisfy them and the integers for the number itself, such as 153=1^3+5^3+3^3. Sample input An input example that satisfies the requirements of the topic.Cases:No sample output 153XxxXxxTopic Analysis:This is a simple problem of brute force cracking. Simply break down the hundred, 10, and digit

Algorithm training 6-1 Recursive two-term coefficient value time limit: 10.0s memory Limit: 256.0MB problem Description Sample Enter an input example that satisfies the requirements of the topic.3 10 Sample output corresponds to the output of the sample input above.Data size and convention the range of each number in the input data.Example: The result does not overflow when int is represented.Topic Analysis:for recursive problems, we pay attention to two points: (1) Find the exit; (2) Find simil

Algorithm training Narcissus time limit: 1.0s memory limit: 256.0MB Narcissus number problem description judge giventhree-digit numberWhether Narcissus number. The so-called Narcissus number refers to its value equal to its own number of cubic meters per digit. Example 153 is a number of daffodils. 153=13+53+33 an integer in the input format. Output format is narcissus number, output "YES", otherwise output "no" (not including quotation marks) sample input 123 sample output no data size and conv

use the sort () function, directly using the Arrays.sort () function array of small to large order, each time the results are directly output.At the same time, I need to clear the results after the loop, I use the Arrays.fill () function, with ' "" to fill, but there are insufficient places, will be emptied after the form of 0 exist, the length of the array has not changed, so need me to further improve, the final evaluation results still have a run timeout!! , for informational purposes onlyIn

; * /*output the entire string in reverse order*/ $ Case 3:Panax Notoginseng for(i = strlen (arr)-1; I >=0; I--) - { theprintf"%c", Arr[i]); + } A Break; the /*converts the corresponding uppercase letter in a string to lowercase letters, lowercase to uppercase*/ + Case 4: - for(i =0; I ) $ { $ if(Arr[i] >='A' Arr[i] 'Z') - { -Arr[i] + = +; the

"Idea": vector perfect solution. Notice the dynamic array of vectors, or else 0."AC Code":#include Blue Bridge Cup ALGO-111 random number (sort, go weight)

Remember:The wording of printf ("%*s%s%*s\n", N-i, "", Arr,n-i, "") is used here,where the * in%*s represents the number of the string s, S is given to the string,This is usually used for printing.AC Code:1#include 2 #defineMAX 503 4 intMainvoid)5 {6 intI,n;7 Chararr[max+1];8scanf"%d",n);9sprintf (arr,"%s","*");Ten for(i =1; I ) One { Aprintf"%*s%s%*s\n", N-i,"", Arr,n-i,""); -sprintf (arr,"%s%s", arr,"**"); - } the - return 0; -}

solution will not be guaranteed. Solution 2: wired The cable solution can only deploy single-mode optical fiber cables at a distance of 12 kilometers, which is more time-consuming and costly. You can say you don't have to think about it. Solution 3: Internet + VPN The Internet can be said to be a public wide area network, and many telecom operators have established a four-way and eight-way long-distance network, which makes it possible for our compan

Algorithm Training S01 StringTime limit: 1.0s memory limit: 256.0MBProblem descriptionS01 string is initially "0"Transform as follows0 Change to 01Input format1 integers (0~19)Output formatS01 string after n-Times transformationSample input3Sample output101Data size and conventions0~19Example code:1 ImportJava.util.Scanner;2 3 Public classMain {4 StaticStringBuffer SB =NewStringBuffer ();5 Public Static voidMain (string[] args) {6Scanner sc =NewScanner (system.in);7 intn =sc.ne

much or not clear, and then on the Internet to learn from the great God code, just understand how.1#include 2#include 3#include 4 using namespacestd;5 structnode{6 ints,e,v;7}edge[100010];8 intm=0, n,val[10010],p;9 voidAddintSintEintv) {Tenedge[++m].s=s; OneEdge[m].e=e; Aedge[m].v=v; - } - intfather[10010]; the intFINDF (inti) {//and check set, intends to use Kruskal to calculate the minimum spanning tree; - if(father[i]==i)returni; - Else returnfather[i]=findf (Father[i]); - } + BO

Algorithm training friendly number time limit: 1.0s memory limit: 256.0MB The problem description has two integers, and if the approximate sum of each integer is equal to the other (except for itself), we call this logarithm friendly. For example:9 Approximate and have: 1+3=44 Approximate and have: 1+2=3So 9 and 4 are not friendly.220 approximate and With: 1 2 4 5 10 11 20 22 44 55 110=284284 Approximate and With: 1 2 4 71 142=220So 220 and 284 are friendly.Write a program to determine whether t

(1, CC, 1): - # Down ins[(rr-1 + i) * m + (j + i)] = (rr-1 + i) * m + (j-1 +i) - forJinchXrange (Rr-2,-1,-1): to # Right +s[(j + i) * m + (cc-1 + i)] = (j + 1 + i) * m + (cc-1 +i) - forJinchXrange (Cc-2,-1,-1): the #Top *S[i * m + (j + i)] = i * m + (j + 1 +)i) $i + = 1Panax Notoginseng returns - the defMultiply (A, b): +n =Len (a) Ac = [] the forIinchxrange (n): + c.append (A[b[i]) - returnC $ $ defDisplacepow (A, k): -

"Ideas": Use of Atoi and itoa, note atoi, three parameters. The first one is the number to convert, the second is the one that is stored in that string, and the third is what the binary is. "AC Code":#include Blue Bridge Cup ALGO-93 inverse number (math, function)