smallest microprocessor

Random increment algorithm (a randomized incremental algorithm)#defineSQR (x) ((x) * (x))#defineEPS 1e-4structp{Doublex, y; P (DoubleXDoubley): x (x), Y (y) {} p (pa, P b): X (b.x-a.x), Y (b.y-a.y) {} p () {} p mid (Pa) { returnP ((a.x+x)/2, (a.y+y)/2); } DoubleCross (P a) { returnx*a.y-y*a.x; } Doublelen2 () {returnSQR (x) +Sqr (y); } DoubleDis (P a) { returnsqrt (SQR (x-a.x) +sqr (ya.y)); } voidprint () {printf ("%f%f\n", x, y); }};structdisc{P o; DoubleR; BOOL

Minimum Inversion numberTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 14879 Accepted Submission (s): 9082Problem DescriptionThe Inversion number of a given number sequence A1, A2, ..., the number of pairs (AI, aj) that SA Tisfy i For a given sequence of numbers a1, A2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we'll Obtain another sequence. There is totally n such sequences as the following:A1, A2, ..., An-1, an

Thinking about it:My train of thought is a bit like sliding window. Assuming that an interval [L, R] has been determined, the sequence from L to R contains exactly the elements of [1, K], then the various positions of the iteration sequence are resumed from R, and if the number of 1 to K is found, the following processing is done:If the number is exactly the number in the L position, then it means that the interval may be shortened, and both L and R are updated to calculate the change in interva

Test instructionsLook for a small number of k in a BST.Ideas:Recursion is more convenient.1 /**2 * Definition for a binary tree node.3 * struct TreeNode {4 * int val;5 * TreeNode *left;6 * TreeNode *right;7 * TreeNode (int x): Val (x), left (null), right (null) {}8 * };9 */Ten classSolution { One Public: A intCountnode (treenode* root)//calculate the number of sub-tree nodes, in fact, you can also directly melt this function into the following function, with K to mark the function. -

. OutputThe minimal length of the rope. The precision should be 10^-2. Sample Input9 12 7 24 9 30 5 41 9 80 7 50 87 22 9 45 1 50 7 0 Sample Output243.06 Sourceasia 1997, Shanghai (Mainland China) Recommendignatius.l Topic Analysis:To find the perimeter of the convex hull. The problem is about the same as getting wall. It is important to note that the number of nodes in this problem may be The code is as

(!Vis[j]) {Vis[j]=true; DFS1 (d+1, cur*Ten+Line[j]); VIS[J]=false; } }}//0 2, 0 3//#define LOCALintMain () {#ifdef LOCAL freopen ("In.txt","R", stdin);#endif intT scanf"%d",T); scanf ("\ n"); while(t--){ Charch; N=0; while(ch = getchar ())! ='\ n'){ if(IsDigit (ch)) {Line[n+ +] = ch-'0'; } } if(n = =2) {printf ("%d\n", ABS (line[0]-line[1]));Continue; } Len1= n>>1; Len2 = nlen1; Ans=1 -; memset (Vis,0,sizeof(VIS)); for(intfi =0; Fi ){ if(Line[f

/*** Enter a set of height between 170 and 190 (5 height), compare height difference, choose the minimum height difference of two height, if the difference in height is the same,* Choose the average height of those two height; from small to large output; Input 170 181 173 186 190 Output 170 173*/public class Mindataheight{public static void Main (string[] args){int a[]={170, 181, 173, 186, 190};System.out.println (Gettarget (a));}public static String Gettarget (int[] a){int lenofa = A.length;int

The main topic: according to the requirements of the topic to construct a sequence, find the shortest sub-sequence, including 1~k.Problem-solving ideas: First, according to the method of the topic constructs the sequence, then uses the Towpointer method, uses the v[i] to record the current [L, R] has several I, when R moves, appears v[i] = = 1 o'clock, C + + (used to record several 1~k numbers); when c = = k, move L, when V [i] = = 0 o'clock, c--。1#include 2#include 3#include 4#include 5 using n

Sample Input1//t3//n0 990 692//adjacency Matrix990 0 179692 179 0Sample Output692Prim1# include 2# include 3# include 4# include 5# include 6# define LLLong Long7 using namespacestd;8 9 Const intinf=0x3f3f3f3f;Ten Const intmaxn=510; One BOOLVIS[MAXN]; A intLOWC[MAXN]; - intN; - intCOST[MAXN][MAXN]; the intMAX =0 ; - - intPrim ()//Dot is 0~n-1 - { + intans=0; -memset (Vis,false,sizeof(Vis)); +vis[0]=true; A for(intI=1; i0][i]; at for(intI=1; i) - { - intMinc=INF; -

Simply to utilize BST's property. The Two-pass code below can be extended to support frequent insert\delete.structsnode{SNode (intrcnt): CNT (rcnt), Pleft (nullptr), Pright (nullptr) {}intCNT; SNode*Pleft; SNode*pright;};classSolution {SNode* Build (TreeNode *p) {if(!p)returnnullptr; SNode*pnew =NewSNode (1); if(p->Left ) {pnew->pleft = Build (p->Left ); Pnew->cnt + = pnew->pleft->CNT; } if(p->Right ) {pnew->pright = Build (p->Right ); Pnew->cnt + = pnew->pright->CNT; } returnpnew;

Title Link: hereSuppose N=B0+B1+B2....+BKIf b2If violence does not have the right result, namely:If b2>n, then n is composed of only two digits in the B-binary.To form a palindrome tree, then Xb+x=n b=n/x-1, (n%x==0,x#include #include#includeusing namespacestd;#definell Long Longll a[ -];BOOLOkintk,ll N) { intCnt=0; while(n) {a[cnt++]=n%K; N/=K; } for(intI=0; i2; i++) { if(a[i]!=a[cnt-i-1])return false; } return true;}intMain () {intT; ll N; scanf ("%d",T); while(t--) {s

Topic Links:2031Http://write.blog.csdn.net/postedit?ref=toolbarTest instructionsIn three-dimensional space, there are n spherical space stations, which give the three-dimensional coordinates x, Y, z and radius of each space station. There may be intersections, inclusions, and other conditions between space stations. If not, the distance between the two space stations is 0. Minimum spanning tree for all space stations.ExercisesThree-dimensional composition the distance between the two space stati

4.3 Given A sorted (increasing order) array with unique integer elements, write a algorithm to create a binary search Tre E with minimal height.This problem gives us an ordered array, let us generate a minimum height of two-fork search tree, in order to achieve the minimum height, it is necessary to fill as far as possible as a full two fork tree, left subtree filled, right subtree as much as possible to fill. and pay attention to the two-fork search tree, left classSolution { Public: TreeNode*

- if(Ans[i +1] >Ans[i]) { -Rank_suffix[i] = n-i +1; + } - //It can be thought that when the numbers are as large as possible, the numbers that are removed from the preceding numbers are larger. + Else if(Ans[i +1] ==Ans[i]) { ARank_suffix[i] = rank_suffix[i +1] +1; at } - //Ans[i + 1] - Else{ -Rank_suffix[i] =1; - } - } in - //divide and conquer, beg K small to intres =1; +

Topic:given the m*n matrix, each position is a non-negative integer, starting from the upper left corner, each time can only go to the right and down, go to the lower right corner, but only to walk once, the sum of the smallest path. Ideas:Because only walk once, so relatively simple, dp[0, 0]=a[0, 0], and dp[x, y] = min (dp[x-1, y] + a[x, y],dp[x, y-1] + a[x, y]).#include Path to minimum sum

) are overwritten. Then the problem becomes the minimum point set of the solution binary graphCovered. The minimum point set coverage of the dichotomy graph = Two The maximum matching of the graph, which is solved by the Hungarian algorithm.There is a place to pay attention to this problem, that is, the number of binary map points. Considering that the original size is 55*55, and the continuous slurry griddo a point, now in order to find out the maximum number of points, assuming that all the mu

" * if B (i,j+1) @paramargs*/ Public Static voidMain (string[] args) {Scanner input=NewScanner (system.in); System.out.println (Enter Number of rows:); intn =Input.nextint (); System.out.println ("Number of input columns:"); intm =Input.nextint (); intA[][] =New int[30] [30]; intB[][] =New int[30] [30]; Char[][]STM =New Char[30] [30]; //Output Matrix for(inti=1;i) { for(intj = 1;j) {A[i][j]= (int) (Math.random () *20+1); B[I][J]=A[i][j]; System.out.printf ("%4d", A

Ask how to find the smallest variable, not the minimum value is how much, the minimum value of the position can also. Like what: $a = 1; $b = 2; $c = 3; Reply to discussion (solution) $_t = Get_defined_vars (); $a =1; $b =2; $c =3; $t = Array_diff_key (Get_defined_vars (), $_t, Array (' _t ' = ")); Echo Array_ Search (min ($t), $t); A The variable name is printed after it gets to that variable function VarName ($v) { $trace = Debug_ba

/* HDU 3157poj 3801 question: a circuit board with N wiring bars (Number 1 ~ N) there are also two power supply terminals +-and then the terminals and the minimum current of M parts are given to find a total current that can make all parts work normally. Then the output impossible is actually a lower bound with a source sink. minimum flow problem handling: 1. construct an additional network 2. find the maximum stream for SS and TT (the full stream of SS and TT has a solution) 3. if there is a so

/*** Question: Enter n integers and output the smallest k of them. * For example, if you enter the 8 numbers 1, 2, 3, 4, 5, 6, 7, and 8, the minimum four digits are 1, 2, 3, and 4. * The simplest way to solve this problem is to sort the input n integers so that the number of K at the top is the minimum number of K. The time complexity of this approach is O (nlogn ). * The question does not require the number of K to be searched, or even the last n-k N