classical webcast

Problem Description: Enter a high-precision positive integer n with the keyboard, and the remaining digits after the s digits are removed to form a new positive integer in the left and right order.Programming given N and S, looking for a scheme to make the remaining numbers oh the smallest number of new numbers.Thinking analysis: Using the greedy method to approximate the target to delete one of the S characters, the total number of each step is to choose a number of the remaining number of the

same probability of being equal to 0.5 of the minimum number, this is the minimum number you want to invite.Ideas:Ask for at least two people birthday the same situation there are many, so we reverse thinking, find out any two people birthdays are not the same probability, when the requirement is to exit the cycle, the result is the minimum number, to pay attention to ISWe want to subtract ourselves here, should be the number of people to be invited, but also to prevent data overflow, so we nee

number and the expected number of gold you'll collect. Errors less than 10-6 'll be ignored.Sample Input31101210 333 6 9Sample OutputCase 1:101Case 2:13Case 3:15Test instructions: Starting from the 1th of the cave, start rolling the dice (six sides), throw the number is how many steps, each step, will wait until the place of the treasure, finally find the treasure of expectationsSuch as:1101Expected =101*1=101210 3Expected =10*1+3*1=1333 6 9Expect =3*1+9*0.5+ (6+9) *0.5=15#include"iostream"#inc

a drink into a meeting point to make a pair, and then the middle must go through the cows that love them.But one problem is that if the simple food-"bull-" drink, then there will be a lot of food and drink by a cow to enjoy the situation, so the cow was torn down, into the food-ox-ox-------the drink, which two cows are the same, the capacity of the side is 1, so that will be avoided, A cow enjoys many pairs of cases, because the ox-to-ox capacity is 1.As for the template, Dinic and SAP are the

#include Non-recursive classical implementations of power functions (time complexity is only logn)

piece is not satisfied, will go back to the first row - } to } + } - the //if you want to place the Queen at the (x, y) point, you need to first determine whether the point can be placed * Public Static BooleanIsOK (intXinty) { $ Panax Notoginseng //Traverse a two-dimensional array to determine whether there are queens in 4 directions - for(inti = 0; i ) { the for(intj = 0; J ) { + //positive hypotenuse x-y fixed va

time from 1A to 1E, from 2a to 2E, we can change it to start from 1b and compare it with 1a first, then compare 2B with 2a, and then compare 1C with the data in the previous group ........ Each time, starting from the Gap Element in the array, it is clear that each element is directly inserted and sorted with the data in its own group. Void shellsort2 (int A [], int N) {Int J, gap; For (Gap = n/2; Gap> 0; Gap/= 2) for (j = gap; j Then, insert the directly inserted sorting part and use the sec

I node is (I-1)/2. Its subnode subscript is 2 * I + 1 and 2 * I + 2 respectively. For example, the subnode subscripts of the first 0th nodes are 1 and 2 respectively. Heap operation-insert and delete The following describes how to create, insert, and delete the smallest heap in "Data Structure C ++ language description Description", and then shows the implementation code. It is best to first clarify the diagram and then look at the code. Heap insertion Each insert operation puts the new data at

Ten beautiful classical sentences1) I love you not because of who you are, but because of who I am when I am with you.I love you not because of who you are, but because of how you feel when you are with you. 2) No man or woman is worth your tears, and the one who is, won't make you cry.No one is worthy of your tears, and those who are worthy of your doing so will not make you cry.　　 3) The worst way to miss someone is to be sitting right beside them k

. println ("Before sorting:"); 13 for (INT I = 0; I Bucket sorting: it is assumed that the input is a real number that is evenly distributed across [0, 1) intervals generated by a random process. Divide the interval [0, 1) into n sub-intervals of the same size (bucket), the size of each barrel 1/N: [0, 1/n), [1/N, 2/N), [2/N, 3/N ),..., [K/N, (k + 1)/n ),... Allocate N input elements to these buckets, sort the elements in the bucket, and connect the bucket to input 0 ≤ A [1 .. n] Sorting of

Classical algorithm learning-fast sorting Quick sorting is the most commonly used algorithm in the interview test. Sorting efficiency is highly efficient among several sorting methods that are the same as O (N * logN). Therefore, this method is often used, and the idea is also the idea of partitioning and recursion. Sample code uploaded to: https://github.com/chenyufeng1991/QuickSort The basic idea of an algorithm is: (1) first extract a number from t

1. There is a soft uniform thin line, in the damping medium for small transverse vibration, the unit length chord resistance $F =-ru_t$. The vibration equation is deduced.Answer: $$\bex \rho u_{tt}=tu_{xx}-ru_t. \eex$$2. The three-dimensional heat conduction equation has the spherical symmetry form $u (x,y,z,t) =u (r,t) $ ($r =\sqrt{x^2+y^2+z^2}$) solution, trial: $$\bex U_t=a^2\sex{u_{rr}+\frac{2u_r}{r}}. \eex$$Proof: by $$\bex U_x=u_r\frac{x}{r},\quad u_{xx}=u_{rr}\frac{x^2}{r^2} +u_r\frac{r-x

0x0001 first group A [1] = 1 0x0001 first group A [2] = 3 0x0011 second group A [3] = 5 0x0101 first group A [4] = 2 0x0010 second group A [5] = 2 0x0010 second group The first group has {1, 1, 5}, and the second group has {3, 2, 3}. Obviously, the two groups can get 5 and 3 by performing the "XOR" solution respectively. After some analysis, I believe the code is not difficult to write. The complete source code is provided below: // Baidu interview question // in the array, all the numbers exce

Classical Sentence Pattern of English Writing) I,~~~ The ~ Est (that) Subject word have ever seen (known/heard/had/read, etc) ~~~ The most adjective noun (that) Main word have ever seen (known/heard/had/read, etc) Example: Helen is the most beautiful girl that I have ever seen.Helen is the most beautiful girl I have ever seen. Mr. Chang is the kindest teacher that I have ever had.Teacher Zhang is the kind teacher I have ever met. 　　 2. Nothing is ~~~

node t4 see:All data are known to have a total of 60 together. The node error cost of node T4 is:Sub-tree error costs are:R (t_t) =\sum{r (i)}= (\frac{2}{5}*\frac{5}{60}) + (\frac{0}{2}*\frac{2}{60}) + (\frac{3}{9}*\frac{9}{60}) =\frac{5}{60 } "width=" 410 "height=" style= "border:0px" >There are 3 leaf nodes on the subtree with T4 as the root node, and finally:A non-leaf node with the lowest alpha value is found, leaving the child null for the left and right.When the alpha value of multiple no

#include Classical algorithms--all combinations of strings

; border:0px; max-width:100%; Height:auto; width:493px ">4 RBM-limited Boltzmann machineRBM 1 Modelscontent=# "style=" >content=# "style=" >content=# "style=" >content=# "style=" >content=# "style=" "> Use formula. Can getresizesmallwidth=832 "class=" En-media "style=" margin:0px; padding:0px; border:0px; max-width:100%; Height:auto; Width:494px ">content=# "style=" >2 Solving the CD algorithmcontent=# "style=" >resizesmallwidth=832 "class=" En-media "style=" margin:0px; padding:0px; borde

for an int array. Please write a heap sorting algorithm. Sorts the array elements.Given an int array a and the size of the array n, return the sorted array.Test examples:[1,2,3,5,2,3],6[1,2,2,3,3,5]Class Heapsort {public:int* heapsort (int* A, int n) { buildmaxheap (A, n);//First build the array A into a large top heap for (int i = n-1; i > = 1; i--) { swap (a[0],a[i]);//Exchange Maxheapfix (A, 0, i-1) with the top maximum of the heap and the last number that is not sort

The simplest coin problem:Title Description:There are 1,5,10,50,100,500 coins each c1,c5,c10,c50,c100,c500, now to use these coins to pay a yuan, the minimum number of coins required.Problem Solving Ideas:Greedy algorithm, unexpectedly may use the face value biggest coin's this greedy strategy to cut.1#include 2 using namespacestd;3 Const intv[6]={1,5,Ten, -, -, -};4 intc[6]={3,2,1,3,0,2};5 intA=620;6 voidsolve () {7 intans=0;8 for(intI=5; i>=0; i--)9 {Ten intT=min (A/v[i],c

-means division, until the user-specified number of clusters. There are two ways to continue dividing clusters according to the SSE selection:(1) Choosing which cluster to divide depends on whether the value of SSE can be minimized by its partitioning. This requires dividing each cluster into two divisions, then calculating the sum of the cluster SSE after the cluster and calculating the difference between its and the binary SSE (of course SSE must fall), and finally selecting the cluster with t