C++

An empty plane, adding a point each time, its coordinates are calculated based on the previous point: (x [I-1] * Ax + Bx) mod Cx, (y [I-1] * Ay + By) mod Cy calculates the square of the distance from the nearest point of a certain set, adding n

After doing this, I found that logarithm is an extremely good thing. At first, I directly used the formula, but it was out of the floating point number range. Later I changed it to the logarithm, But it timed out. I also struggled. Later I found

  Problem: calculate all the intersection points composed of n straight lines. Analysis: the first three straight lines are well analyzed and easy to understand. Therefore, we can add one to the fourth line from the third to analyze all the

DescriptionHardwoods are the botanical group of trees that have broad leaves, produce a fruit or nut, and generally go dormant in the winter.America's temperate climates produce forests with hundreds of hardwood species -- trees that share certain

First, we need to deal with the input. The general idea of solving the problem is to rearrange the given c array and r array according to the rank of each historical event, that is, the number of the earliest event is placed in the first place of

#include#include#include#include#include#include#include#include#includeusing namespace std;const int maxn = 115;const int inf = 9999999;char mat[ maxn ][ maxn ];int vis[ maxn ][ maxn ][ 2 ][ 2 ];const int dx[]={1,-1,0,0};const int

Analysis: although we can see that there is a directed edge, we can do it with a undirected edge!   #include #include #include using namespace std; const int maxn=6010; struct node{ int v; node *next; }tree[maxnv]){

Link: HDU 2196 Computer Analysis: first, obtain the maximum distance from any point to another point, and then update its neighbor point with this point as the center, Use the updated vertex to update the adjacent vertex ...... so on! Code:   #

Warm upTime Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)Total Submission (s): 2012 Accepted Submission (s): 474   Problem Description  N planets are connected by M bidirectional channels that allow instant

This is a simple graph topic. You only need to use the width-first search (BFS) to mark the distance between nodes. My problem-solving code is as follows:   # Include # include # include # include # include # include # include # include

Just like a previous question, it is always a problem of interval dyeing. I still use my old method to implement interval discretization + binary interval endpoint + interval processing, and the time is quite short.        The problem is that the

Shortest PathTime Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission (s): 22174 Accepted Submission (s): 9436   Problem DescriptionIn each year's competition, all the finalists will get a very beautiful

Problem DPrince and PrincessInput: Standard Input Output: Standard Output Time Limit: 3 Seconds In an n x n chessboard, Prince and Princess plays a game. The squares in the chessboard are numbered 1, 2, 3... n * n, as shown below:   Prince stands

In the w array, the value is 1. The array is initially 0. How many sort methods are available to make the array Lucky. The total number of solutions should be provided after each update.   Dynamic Planning can be used to obtain the result after the

Question: There are some small islands with some edges that allow you to add one edge to minimize the number of bridges on those sides. Practice:1. Map the island as a vertex and the island connected to the edge. 2. Find all the strongly connected

#include #include #include int main(){ int num[1000],maxnum; char word_list[1000][100]; char excuse[1000][100]; char s[1000][100]; int i,j,k,l; int cas=1; int n,m; while(scanf("%d %d",&m,&n)!=EOF) { for(i=0;imaxnum) maxnum=num[j]; } }

// TIM1 frequency division # defineTIM1_DIV1 (1-1) # defineTIM1_DIV2 (2-1) # defineTIM1_DIV4 (4-1) # defineTIM1_DIV8 (8-1) # defineTIM1_DIV9 (9-1) # defineTIM1_DIV18 (18-1) # defineTIM1_DIV72 (72-1) # defineTIM1PinA_Enb TIM1-> CCER | = 0X0001 //

Idea: weighted and query setAnalysis:1. Given n conditions, we need to find the first number that does not meet the conditions.2 each condition gives the parity of 1 in the range [l, r]. Obviously, the number of 1 in the range [l, r] can be [0, r]-[0

RectanglesTime Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission (s): 11631 Accepted Submission (s): 3716   Problem DescriptionGiven two rectangles and the coordinates of two points on the diagonals of each

Add a vis Array Based on the template of my minimum spanning tree Prim algorithm to identify whether a node has been added to set T. Here, the min_dis of a node cannot be used as whether to add the node to T. Because the connected edge is given in