If I understand the problem correctly, I would ask a user what operation he wants to go with: math_option =raw_input(‘Choose the math operation: ‘) and later on check what option was chosen: if math_option == “add”: print add(x, y) if math_option == “multiply”: add_num = add(x,y) mul_num = multiply(math_option,z) print mul_num and so on 1 … Read more
Do like this: public int formulaMethod(int n) { int result = 1; for (int i = 1; i <= n; i++) { result *= i; } return result; } Hope it helps 0 solved Java Math Formula Loops
In i = i– – –i you have: i–, a post-decrement, which retrieves the current value of i (1) and then decrements i to 0 – –i, a pre-decrement, which decrements i again and retrieves the updated value, -1 So you end up with i = 1 – -1 which is 2. Needless to say, … Read more
I being a number, there is 10 possibilities. S being a character, there is 26 possibilities. The total number of combinations is 10 power (TheNumberOfI) * 26 power (TheNumberOfS) This can be dynamically solved using a simple function that counts the number of S and the number of I and uses the results in the … Read more
Here’s the problem: double math = 1/k; and formula2 -= 1/k; k is an int variable, so the JVM won’t never return a decimal number in this statement. It will take only two possible values: 0 (if k > 1) or 1 (if k == 1) because the JVM performs the division before promoting the … Read more
The first character in an identifier must be a-z, A-Z, or underscore, for 53 possibilities. Subsequent characters can be digits 0-9 too, for 63 possibilities. So n characters give 53 * (63**(n-1)) legal identifiers: Length Names 1 53 2 3339 3 210357 // much less than 1000000 4 13252491 // much more than 1000000 Some … Read more
This is because both the following operations are different : 5 * M_PI / 4 and 5 / 4 * M_PI The operations are performed left to right. On the first line, 5 * M_PI (int * float) returns a float (15.7…) which is divided by an int (float / int) returning a float (3.9…). … Read more
Compare X ≡ s1 * m2 with Y ≡ s2 * m1. If X > Y, then s1 / m1 > s2 / m2. No division is required to do the comparison. The caveat to this solution is that s1, s2, m1, and m2 should all have the same sign, and m1 and m2 should … Read more
var x = 5; // x = 5 x *= 2; // multiply x with 2 (x = 10) console.log(++x); // console.log x plus 1 (11) A more common way of using this syntax is with plus or minus: x += 1; // is a shorthand for x = x + 1; x *= 2; … Read more
It always helps if you include the error. There are two ways to fix that error: Interpolate the value: puts “you would like #{number} much better” Turn it from a number to a string: puts “you would like ” + number.to_s + ‘much better’ The former, #{…} syntax, evaluates the content of the braces as … Read more
var lstPnts = new List<PointF>(); lstPnts.Add(new PointF(23.92f, 6)); lstPnts.Add(new PointF(23.88f, 0)); lstPnts.Add(new PointF(0, 0)); lstPnts.Add(new PointF(0, 6)); var avgPoint = new PointF(lstPnts.Average(t=>t.X),lstPnts.Average(t=>t.Y)); var ordered = lstPnts.OrderBy(t => Math.Atan2(avgPoint.Y – t.Y, avgPoint.X – t.X)).ToArray(); We found an average point. Then we calculated the angle between mid point and the other points and sorted our array with … Read more
At the end of the for loop s and k are equal. Before the next iteration k != n is checked. This is equivalent to s != n. So the loop runs until s == n holds and then n is returned. So the function get the input n, runs for some time and returns … Read more
The solution is : for n=1: = 1 for n=2: = 1+2 => 1+2+3 => 1-2+3 for n=3: = 1+2 for n=4: = 1+2+3 => -1+2+3 for n=5: = 1+2+3 => 1+2+3+4-5 for n=6: = 1+2+3 for n=7: = 1+2+3+4=>1+2+3+4+5=>1+2+3-4+5 for any n, first calculate the S(k)=1+2+3+…+k where S(k)>n and x=S(k)-n is an even number. … Read more
Use sine rule:http://media.tiscali.co.uk/images/feeds/hutchinson/ency/c02395.jpg If you have side a and angle A, and you want to work out the length of side b and you have angle b, the formula would be b = sin B * (a/sinA) In Javascript this can be done by var B = 45; //size of angle B var a = … Read more
The watt to dBm conversion follows the following rule: dBW = 10 * log10(P[w]) dBm = 10 * log10(1000 * P[w]) = 10 * log10(P[w]) + 10 * log10(1000) = 10 * log10(P[w]) + 10 * 3 = 10 * log10(P[w]) + 30 = dBW + 30 Hence, the inverse path will be: P[w] = … Read more