math

问题 I am trying to write a function that will randomly return an (x,y) co-ordinates around a given circumference so if I have a point that's at (0,0) (being the center of the div) how can I write a function that randomly places other entities that appear among the outer edge of a circle. All I need is the equation i know it has something to do with getting the distance from the center to the circumference edge just no idea how to calculate it and randomize it so it looks good. 回答1: Just get a

问题 Math.atan2 uses a convention for angle that starts at -pi/2 at right, increasing til pi/2 at the same point. The usual angle notation we see in most math works starts at 0 at right, and counter-clockwise increases til pi*2 at the same point. usual Math.atan2(y,x) right 0/2*pi -2/2*pi top 1/2*pi -1/2*pi left 2/2*pi 0 down 3/2*pi 1/2*pi This is not just unusual, but inconsistent with the other trigonometric functions. Why? Is there any problem if I define my own atan2 that works the way I'm

问题 Math.atan2 uses a convention for angle that starts at -pi/2 at right, increasing til pi/2 at the same point. The usual angle notation we see in most math works starts at 0 at right, and counter-clockwise increases til pi*2 at the same point. usual Math.atan2(y,x) right 0/2*pi -2/2*pi top 1/2*pi -1/2*pi left 2/2*pi 0 down 3/2*pi 1/2*pi This is not just unusual, but inconsistent with the other trigonometric functions. Why? Is there any problem if I define my own atan2 that works the way I'm

问题 I am trying to write a code that find the sum of all the possible numbers in a 2d matrix. But the catch is that you must choose one element from one row. #include <algorithm> #include <iostream> #include <vector> #include <climits> #include <math.h> using namespace std; int main() { int n; cin>>n; int array[n][n]={0}; for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ cin>>array[i][j]; } } int power=pow(n,n); int sum[power]={0}; for(int i=0;i<power;i++){ for(int j=0;j<n;j++){ for(int l=0;l<n;l++){

问题 I am trying to write a code that find the sum of all the possible numbers in a 2d matrix. But the catch is that you must choose one element from one row. #include <algorithm> #include <iostream> #include <vector> #include <climits> #include <math.h> using namespace std; int main() { int n; cin>>n; int array[n][n]={0}; for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ cin>>array[i][j]; } } int power=pow(n,n); int sum[power]={0}; for(int i=0;i<power;i++){ for(int j=0;j<n;j++){ for(int l=0;l<n;l++){

问题 I am trying to write a code that find the sum of all the possible numbers in a 2d matrix. But the catch is that you must choose one element from one row. #include <algorithm> #include <iostream> #include <vector> #include <climits> #include <math.h> using namespace std; int main() { int n; cin>>n; int array[n][n]={0}; for(int i=0;i<n;i++){ for(int j=0;j<n;j++){ cin>>array[i][j]; } } int power=pow(n,n); int sum[power]={0}; for(int i=0;i<power;i++){ for(int j=0;j<n;j++){ for(int l=0;l<n;l++){

问题 I encountered a hard question I don't know the answer to: "Rearrange the digits from an integer in blocks of two with a recursive function" here's an example: Input: 123456 unsigned long pairinvPrint(unsigned long number) { printf("%d", number % 100); if ((number / 100) <= 99) { printf("%d", number / 100); } else { pairinv(number / 100); } } Output: 563412 More I/O Examples: 42 -> 42; 1234 -> 3412 However, the set circumstances to do this are hard (no loops, arrays, pointers, global- or

问题 Does an equivalent of Java RoundingMode.HALF_DOWN exist in C#? For example, I want to round 1.265 to 1.26 , and 1.266 to 1.27 . If not, is there a simple way to do it? 回答1: Have a look at Math.Round e.g. double[] tests = new double[] { 1.265, 1.266, }; var demo = tests .Select(x => $"{x} -> {Math.Round(x, 2, MidpointRounding.AwayFromZero)}"); var report = string.Join(Environment.NewLine, demo); Console.Write(report); Outcome: 1.265 -> 1.26 1.266 -> 1.27 回答2: Use the .Round method with the

问题 Does an equivalent of Java RoundingMode.HALF_DOWN exist in C#? For example, I want to round 1.265 to 1.26 , and 1.266 to 1.27 . If not, is there a simple way to do it? 回答1: Have a look at Math.Round e.g. double[] tests = new double[] { 1.265, 1.266, }; var demo = tests .Select(x => $"{x} -> {Math.Round(x, 2, MidpointRounding.AwayFromZero)}"); var report = string.Join(Environment.NewLine, demo); Console.Write(report); Outcome: 1.265 -> 1.26 1.266 -> 1.27 回答2: Use the .Round method with the

问题 Hey, I'm writing a formula with three indexes i,j,k. At the end of the line I'd like to put this: i=1,...,a j=1,...,b k=1,...,n But I'd like it in smaller font and stacked above each other. Can someone tell me a command which can accomplish this? \mbox can't do math mode, I think. 回答1: Try the \substack command: z_i = a_j + b_k \qquad \substack{ i=1,\dots,a \\ j=1,\dots,b \\ k=1,\dots,n} 来源： https://stackoverflow.com/questions/5009782/latex-stack-three-lines-in-math-mode