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Netflow addition with ELK stack
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Behaviour of Array addition with labels
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White Noise kernel addition with grid interpolation
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Rationals do not support addition with empty set
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Support for ECC in addition with RSA.
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Problem in addition with htitle (can't scroll to last tabs)
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double view of IE's and KH's in addition with to long numbers if come from Nightscout/AAPS
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Mismatched types with addition
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License addition
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Update README with Style Addition
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Update README with Sender addition
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Binary search with addition and subtraction
20170210 09:46:48最近在学习Algorithms 4th, 做个... Binary search with only addition and subtraction. [Mihai Patrascu] Write a program that, given an array of N distinct int values in ascending order, determines whether a最近在学习Algorithms 4th, 做个记录
题目
Binary search with only addition and subtraction. [Mihai Patrascu] Write a program that, given an array of N distinct int values in ascending order, determines whether a given integer is in the array. You may use only additions and subtractions and a constant amount of extra memory. The running time of your program should be proportional to log N in the worst case.
仅用加减实现的二分查找。编写一个程序给定一个含有N个不同int值并按照升序排序的数组，判断是否有给定的整数，只能用加法和减法以及额外的内存，程序运行时间在最坏情况下应该和logN成正比解决方法
package day1; import edu.princeton.cs.algs4.*; import java.util.Arrays; import java.util.HashSet; public class Mihai_Patrascu { public static void main(String[] args) { // TODO Autogenerated method stub int[] arr = getArr(20000); StdOut.println(); Arrays.sort(arr); // 排序 for (int i :arr){ StdOut.print(i+" "); } // fbnq(10, 5, 2); int index = max(arr, 0, arr.length, 10); StdOut.println("answer: "+index); } /** * * @param arr 数组 * @param lo 起始下标 * @param hi 最终下标 * @param goal 查找的参数 * @return */ public static int max(int[] arr, int lo, int hi, int goal) { if (lo > hi) return 1; // init if (hi  lo <= 2) { if (arr[lo] == goal) return lo; if (arr[hi] == goal) return hi; else { return 1; } }else { int k3 = 0; int i = 3; while (k3 < hi) { i++; k3 = fbnq(i, lo, 1)  lo; StdOut.print(k3 + " "); } StdOut.println(); int k2 = fbnq(i  1, 0, 1) + lo; if (arr[k2] > goal) return max(arr, lo, k2  1, goal); if (arr[k2] < goal) return max(arr, k2 + 1, hi, goal); else return k2; } } // 斐波那契数列 /** * * @param n 下标 * @param first 第一个数 * @param second 第二个数 * @return n小标下的数字 */ public static int fbnq(int n,int first, int second) { int a = first; int b = second; if (n < 3) return 1; else { int i = 3; while (i <= n) { int temp ; temp = a; a = b; b = temp + a; i++; // StdOut.print(b+" "); } return b; } } // 创建不重复的数组 public static int[] getArr(int arrSize) { HashSet< Integer> set = new HashSet<>(); int[] arr = new int[arrSize]; for (int i = 0; i<arrSize; i++){ int a = StdRandom.uniform(arrSize * 2); // 可能会再创建的时候多费一点时间, 但是有利于查找到对应的数字 int count = set.size(); set.add(a); if (count < set.size()) { arr[i] = a; }else { i; } } return arr; } }
最后
stakoverflow上有关于斐波那契search和binary search谁快的问题，答者扯到的硬件方面不是很了解
附上链接
也没怎么精简校验，有错误可以联系我，会及时纠正 
CMD + Click selection addition removed with this package
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Limit with concurrent document addition?
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Improvements to ISystem with Entity Addition/Removal
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Xdrip with libre raw value in addition
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AD Kibana IntegTest addition to build workflows WITH Security
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Addition of Parquet Output Type with Fixed Author
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make usbmount play nicely with systemd with the addition of a systemd…
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Possible addition to new `with` syntax.
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Make execTransactionIfApproved handling of failed transactions in team addition consistent with ...
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algorithm 1.4.22 Binary search with only addition and subtraction
20161118 16:01:48题目：Binary search with only addition and subtraction. [Mihai Patrascu] Write a program that, given an array of N distinct int values in ascending order, determines whether a given integer is ...题目：Binary search with only addition and subtraction. [Mihai Patrascu] Write a program that, given an array of N distinct int values in ascending order, determines whether a given integer is in the array. You may use only additions and subtractions
and a constant amount of extra memory. The running time of your program should be proportional to log N in the worst case
算法：
package com.frozenxia.algorithm.basic; public class FibonacciSearch { public int min(int a, int b) { return a < b ? a : b; } public int search(int[] arr, int x, int n) { /* Initialize fibonacci numbers */ int fibMMm2 = 0; // (m2)'th Fibonacci No. int fibMMm1 = 1; // (m1)'th Fibonacci No. int fibM = fibMMm2 + fibMMm1; // m'th Fibonacci /* * fibM is going to store the smallest Fibonacci Number greater than or * equal to n */ while (fibM < n) { fibMMm2 = fibMMm1; fibMMm1 = fibM; fibM = fibMMm2 + fibMMm1; } // Marks the eliminated range from front int offset = 1; /* * while there are elements to be inspected. Note that we compare * arr[fibMm2] with x. When fibM becomes 1, fibMm2 becomes 0 */ while (fibM > 1) { // Check if fibMm2 is a valid location int i = min(offset + fibMMm2, n  1); /* * If x is greater than the value at index fibMm2, cut the subarray * array from offset to i */ if (arr[i] < x) { fibM = fibMMm1; fibMMm1 = fibMMm2; fibMMm2 = fibM  fibMMm1; offset = i; } /* * If x is greater than the value at index fibMm2, cut the subarray * after i+1 */ else if (arr[i] > x) { fibM = fibMMm2; fibMMm1 = fibMMm1  fibMMm2; fibMMm2 = fibM  fibMMm1; } /* element found. return index */ else return i; } /* comparing the last element with x */ if (fibMMm1 != 0 && arr[offset + 1] == x) return offset + 1; /* element not found. return 1 */ return 1; } int fbsearch(int[] arry, int target) { int fm1 = 1; int fm2 = 0; int fm0 = fm1 + fm2; while (fm0 < arry.length) { fm2 = fm1; fm1 = fm0; fm0 = fm1 + fm2; } int offset = 1; while (fm1 > 0) { int mid = min((offset + fm1), arry.length  1); if (arry[mid] == target) return mid; else if (arry[mid] < target) { fm0 = fm2; fm1 = fm1  fm2; fm2 = fm2  fm1; offset = mid; } else { fm0 = fm1; fm1 = fm2; fm2 = fm0  fm1; } } return 1; } public static void main(String[] args) { int arry[] = new int[100]; for (int i = 0; i < 100; i++) { arry[i] = i + 8; } FibonacciSearch fs = new FibonacciSearch(); for (int i : arry) { // System.out.println(fs.search(arry, i)); // System.out.println(fs.search(arry, i, arry.length)); System.out.println(fs.fbsearch(arry, i)); } // System.out.println(fs.fbsearch(arry, 1887)); } }
Reference：http://www.geeksforgeeks.org/fibonaccisearch/

Removed float addition in simulator
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Fix error when clone item with addition datatime fields
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Added targeting support with the addition of the TargetingFilter.
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Addition XBian tab with xbian settings in default XBMC webinterface
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This is because with the addition of “App Resource Library
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Addition
20140820 11:41:19Six years ago, a robot, Bob, with infant's intelligence has been invented by an evil scientist, Alice. Now the robot is six years old and studies in primary school. Addition is the first oProblem
Six years ago, a robot, Bob, with infant's intelligence has been invented by an evil scientist, Alice.
Now the robot is six years old and studies in primary school. Addition is the first operation he learned in math. Due to his strong reasoning ability, he could now conclude a+b=12 from a=2 and b=10.
Alice wanted to test Bob's addition skills. Some equations were given to Bob in form of a=2, b=10, c=4, and Bob has to find out the answers of questions like a+b, a+c, etc.
Alice checked Bob's answers one by one in the test papers, and no mistake has been found so far, but Alice lost the given equations after a cup of coffee poured on them. However she has some of Bob's correct answers, e.g. a+b=12, a+c=6, c+d=5. She wants to continue with the checkable equations, e.g. b+d=11 could be concluded by a+b=12, a+c=6, c+d=5, and thus the question b+d is checkable.
To prevent the artificial intelligence technology from being under the control of Alice, you disguised yourself as her assistant. Now Alice wants you to figure out which of the rest of questions are checkable and their answers.
Input
The first line of the input gives the number of test cases, T. T test cases follow.
The first line of each test case contains a single integer N: the number of correctly answered questions. Each of the next N lines contain one correctly answered question in the form "x+y=z", where x and y are names of variables and z is a decimal integer.
The next line contains a single integer Q: the number of remaining questions. Each of the next Q lines contain one question in the form "x+y", where x and y are names of variables.
Output
For each test case, the first line of output contains "Case #x:", where x is the test case number (starting from 1). For each question in the input that was checkable, output a single line with the answer in the form "x+y=z", where x and y are names of variables andz is a decimal integer. Questions should be listed in the same order as they were given in the input. Please do NOT ignore duplicated questions, since Alice would fire you if you pointed any mistake of hers.
Limits
Names of variables are strings of lowercase English letters. Each name contains at most 10 characters.
200000 ≤ z ≤ 200000
There is no contradiction in the answered questions.
Small dataset
T ≤ 10
N ≤ 10
Q ≤ 10
Large dataset
T ≤ 3
N ≤ 5000
Q ≤ 5000
Sample
Input
Output
2 2 apple+banana=10 coconut+coconut=12 5 apple+banana apple+banana apple+apple banana+apple peach+apple 3 a+b=3 b+c=3 c+d=3 4 a+c a+d b+c b+d
Case #1: apple+banana=10 apple+banana=10 banana+apple=10 Case #2: a+d=3 b+c=3
将变量抽象为图顶点，等式抽象为图的边，看能否有奇数条边的通路即可。使用BFS搜索。
#include <iostream> #include <fstream> #include <vector> #include <string> #include <unordered_map> #include <queue> #include <sstream> using namespace std; struct GEdge { int To; // the other end int Value; // value of the edge GEdge(int to, int val) : To(to), Value(val) {} }; typedef vector<GEdge> EdgeArray; vector<EdgeArray> g_Graph; // all edges from node i unordered_map<string, int> g_Nodes; // all nodes int g_NodeCount; // used to generate the node index void add_edge(int node1, int node2, int val) { // undirected graph g_Graph[node1].push_back(GEdge(node2, val)); g_Graph[node2].push_back(GEdge(node1, val)); } int get_node(const string &name) // fetch or create a node { if (g_Nodes.count(name) > 0) return g_Nodes[name]; return (g_Nodes[name] = g_NodeCount++); } bool bfs_search_path(int &val, int from, int to) { vector<bool> visited(g_NodeCount, false); queue<GEdge> Q; for (auto edge : g_Graph[from]) Q.push(edge); while (!Q.empty()) { GEdge cur = Q.front(); Q.pop(); int node = cur.To; if (node == to) { val = cur.Value; return true; } for (auto e : g_Graph[node]) { for (auto ee : g_Graph[e.To]) // two loops to find odd edges { if (visited[ee.To]) continue; visited[ee.To] = true; Q.push(GEdge(ee.To, cur.Value  e.Value + ee.Value)); } } } return false; } void work_single(istream &input, ostream &output) { int N; input >> N; // clear graph g_Graph.clear(); g_Nodes.clear(); g_NodeCount = 0; g_Graph.resize(N * 2); // read and build graph for (int i = 0; i < N; i++) { string line; input >> line; // find +,= position size_t j; for (j = 0; j < line.length(); j++) { if (line[j] == '+'  line[j] == '=') line[j] = ' '; } string name1, name2; int val; istringstream line_stream(line); line_stream >> name1 >> name2 >> val; // add edge int node1 = get_node(name1); int node2 = get_node(name2); add_edge(node1, node2, val); } int Q; input >> Q; for (int i = 0; i < Q; i++) { string line; input >> line; // find + size_t j; for (j = 0; j < line.length(); j++) { if (line[j] == '+') break; } string name1 = line.substr(0, j); string name2 = line.substr(j + 1, line.length()  j  1); if (g_Nodes.count(name1) == 0  g_Nodes.count(name2) == 0) continue; int node1 = get_node(name1); int node2 = get_node(name2); int val; if (bfs_search_path(val, node1, node2)) output << line << "=" << val << endl; } } void work(istream &input, ostream &output) { int case_count; input >> case_count; for (int i = 0; i < case_count; i++) { output << "Case #" << i + 1 << ":" << endl; work_single(input, output); } } int main(int argc, char *argv[]) { bool use_stdio = false; if (use_stdio) work(cin, cout); else { ifstream input("Clargepractice.in"); ofstream output("Clargepractice.out"); work(input, output); } return 0; }

Addition of MOG2/KNN constructor with options
20210109 19:16:28<div><p>Hi guys! this is my first contribution to this project and I wanted to ensure that this code is correct in terms of the contributions guideline. So a little hand holding would be appreciated. ...