arria nlg

it is necessary to use a double argument on the case. 3.2-1Show that if f (n) and g (n) is monotonically increasing functions, then so is the functions f (n) + (g (n) and F (g ()), and If f (n) and g (n) is in addition nonnegative, then f (n) • g (n) is monotonically increasing. High school math problem, maybe a junior high school math problem? 3.2-2Prove equation (3.16). a^logb(c) = c^ logb(a) 3.16 Go on 3.2-3Prove equation (3.19). Also prove that and n! = ω(2^n)n! = o(n^n) lg(n!) = Θ

counter array instead of a bit array to support deletion. A more important question is how to determine the size of the bit array m and the number of hash functions based on the number n of the input elements. The error rate is minimal when the number of hash functions k= (LN2) * (m/n). In cases where the error rate is not greater than E, M is at least equal to N*LG (1/e) to represent a collection of any n elements. But M should be bigger, because it also has to ensure that at least half of th

counter array instead of a bit array to support deletion. A more important question is how to determine the size of the bit array m and the number of hash functions based on the number n of the input elements. The error rate is minimal when the number of hash functions k= (LN2) * (m/n). In cases where the error rate is not greater than E, M is at least equal to N*LG (1/e) to represent a collection of any n elements. But M should be bigger, because it also has to ensure that at least half of th

be GT;=NLG (1/e) *lge probably NLG (1/e) 1.44 times times (LG says 2 logarithm). For example, if we assume that the error rate is 0.01, then M should be about 13 times times that of N. So k is probably 8. Note that here m is different from the unit of N, and M is a bit, and N is in the number of elements (exactly the number of different elements). Usually the length of a single element has a lot of bit. So

is counting Bloom filter, which can support deletion by replacing the bit array with a counter array. There is a more important question, how to determine the size of the bit array m and the number of hash functions according to the number of input elements N. The error rate is minimized when the number of hash functions is k= (LN2) * (m/n). In cases where the error rate is not greater than E, M must be at least equal to N*LG (1/e) to represent a collection of any n elements. But M should also

same, the coefficient of merge sorting is larger than the fast one.Q: What are the improvements for merge sorting? A: The insertion sort is used when the array length is k, because the insertion sort is appropriate for sorting the decimal group. Introduced in the introduction of algorithms study Questions 2-1. The complexity is O (NK+NLG (n/k)), when K=o (LGN), the complexity is O (NLGN) Five, fast sequencing Tony Hoare was invented in 1962, known as

Title: The maximum and the number of sub-arrays of an array, which requires O (n) time complexity.Because of the O (n) time complexity limit, the O (n^2) method of brute force solution is certainly not. Consider recursively the maximum of the subarray of an array a[n], which can be decomposed to the maximum of the a[i] sub-array, and to a condition between a[n-i-1]. A[n] The maximum sum of the subarray maximum and equal to the A[i] sub-array; A[n] Sub-array max and equals a[n-i-1];

The length is n, divided into n/k decimal group, each length is KAsk k what value to make the most efficientEasy to get O (n) =nk+nlg (n/k)NK is n/k a decimal group to insert sort kxk, get NKNLG (n/k) is the n/k of the number of clusters need to merge LG (N/K), each time the cost of NAnswershould be a LGNO (n) =nk+nlgn-nlgkIf K>LGN, then the first half is greater than the merge sort time complexity, so k maximum is LGN.Maximum value should be the opt

[]= {19,3,12,4,7,6,12,9,23,14,29,13,34,8}; Debug (A,sizeof (a)/sizeof (int), 0); Insertionsort (A,sizeof (a)/sizeof (int)); MergeSort (A,0,sizeof (a)/sizeof (int)-1); Debug (A,sizeof (a)/sizeof (int), 0); } static void Insertionsort (int* a,int N) {int J; for (j=1;j Although the worst-case runtime for the merge sort is theta (N*LGN), the worst-case runtime for the insert sort is θ (n squared), but the constant factor in the insert sort makes it run faster when n is smaller. Therefore

and R two. Initial: K=p,a[p...p-1] is empty, so it is sorted and established. Hold: Before the K iterations, A[p...k-1] is sorted, and because L[i] and r[j] are the smallest of the remaining elements in L and R, so only the smallest elements in l[i] and r[j] can be placed in a[k], and k+1] is sorted before the A[P...K iterations, and L[i] and r[j] are the smallest remaining two elements. Termination: K=q+1, and A[P...Q] is sorted, and that's what we want, so the proof is over.examples of merge

division of labor to d) Step 1: Put 1, 1 + K, 1 + 2 k, 1 + 3 K ,......; 2, 2 + k, 2 + 2 K, 2 + 3 K ,......;...... As a separate set Step 2: Sort each set at O (NLG (N/K) e) Step 1: Same d) Step 1 Step 2: Merge K paths using heap sorting8-6 merge lower bounds of sorted lists A) 2N number. N numbers are randomly selected. The optional methods include C (2n, n) B) 2 ^ h> = C (2n, n) ====> h> = lg (2n )!) -2lg (N !) ====> H >=2nlg2n-2 nlgn ====> H >=2nlg

Poj2299 -- merge sort to calculate the number of reverse orders, poj2299 -- reverse order /** \ Brief poj2299 ** \ param date 2014/8/5 * \ param state AC * \ return memory 4640 K time 3250 ms **/# include UseMerge SortingSequentialNumber of reverse orders Poj2299 Merge (merge) sort the number in reverse order, correct the error why sum is always 0; help fuel, coming soon to the provincial Competition This assignment is incorrect.For (int j = 0; j Right [j] = s [j + ls];High scores and answers

. When the error rate p is not greater than E: Launch: If the error rate is not greater than E, m must at least be equal to the set of any n elements. However, m should be larger, because it must ensure that at least half of the bit array is 0, then m should be greater than or equal to approximately 1.44 times that of nlg (1/E. The mathematical principle behind the bloom filter is that the probability of two completely random mathematical conflicting

once. The expected running time of this algorithm is O (NK + NLG (N/K )). How to Select K in practice. Analysis: (1) the running time of the algorithm consists of two parts: the first is the fast sorting time, but the insert sorting time. The only difference between the former and the standard quick sorting time is that the depth of recursion is relatively low. Refer to the recursive tree used to analyze the algorithm execution time, the height of th

C language qsort function algorithm performance test, qsort performance test An intuitive method to measure the complexity of an algorithm is to measure the running time of an algorithm of a certain magnitude of data. Taking the qsort provided by C language as an example, we can test the computing time of the qsort with 1 million of the data volume to detect the time cost of O (nlg (n: The C code is as follows: # Include Compile and run the SD

http://www.lydsy.com/JudgeOnline/problem.php?id=3319I think it is a template problem to review the next hld ....... ...........Then nlg^2n was taken by the tle ......... .............Then look at the Qaq,,, this ...God-Problem practice ... I'll write again later ...HLD of Tle:#include 　　 DescriptionGiven a tree, the color of the edges is black or white, and the original is all white. Maintain two operations:1. Query the label of the first black

ExercisesIn fact, is to seek three-dimensional partial order the longest chain. Similar to the three-dimensional inverse pairs, we can use a tree-like array to set up a balanced tree to achieve.DP equation: F[i]=max (f[j]+1) a[j]We sort by one-dimensional, another creates a tree-like array, and inserts the third dimension into each node of the tree-like array.In addition to the weights, each node also maintains an MX that represents the largest f[i in the subtree].So we can

Problem definition:Suppose A[1...N] is an array with n different numbers. If IFor example arrays To solve this problem intuitively, using the brute force solution, the time complexity is O (n^2) for every number num to traverse all the numbers in the array after Num.The implementation code is as follows:Another way, is to use the partition method, the entire array is first divided into two parts, and then, the number of two subgroups to solve the number of reverse order, when two sub-arrays have

1 //Max_heap and Priority queue2 //The index must be [1,size],not [0,size-1]3#include 4 using namespacestd;5 intarr[ -] = {1,4,2,3,9,7,8, -,Ten, -};6 #definePARENT (i) (I/2)7 #defineLeft (i) (i*2)8 #defineRight (i) (i*2+1)9 #defineERROR (-1)Ten //dump an array One voidDumpintArr[],intCNT) A { - for(intI=0; i){ -cout" "; the } -coutEndl; - } - + //ensure the Max heap time=o (LG (n)) - voidMax_heapify (intArr[],Const intSizeConst intindex) + { A intL=left (index), R=right (index), la

Sorting refers to arranging the collection of elements in the specified order. There are usually two sorts of sorting methods, ascending and descending. For example, the integer set {5,2,7,1} is sorted in ascending order, the result is {1,2,5,7}, and the result is {7,5,2,1} in descending order. In general, the purpose of sorting is to make the data appear in a more meaningful way. Although the most significant sort of application is arranging data to show it, it can often be used to solve other