Instead of using Math.floor(Math.random() * 100) twice, use it only once because every time it will generate a new number and assign it to a variable & check if that is between 50 & 60. The result of Math.floor(Math.random() * 100) inside console.log(); & if () is very unlikely to be equal. So even if … Read more
No, in fact this line of code outputs 0: Math.floor(0.9); Math.floor always rounds to the nearest whole number that is smaller than your input. You might confuse it with Math.round, that rounds to nearest whole number. This is why this code always outputs 1 or 0, since input never gets to 2 or bigger: Math.floor(Math.random()*2) … Read more
Denote A[i] – the i’th digit of A. We first need to understand that to get N+M=10^X-1, we need N[i]+M[i]=9 for all i. Since M[i]=N[X-i], it means we need N[i] + N[X-i] = 9. This means, once N[i] is set, also N[X-i]. We can now derive a recursive formula: F(X) = 10*F(X-2) The idea is … Read more
Building on what @Aditi Rawat said, I’ve wrote a little program that does what you need using the quadratic roots formula. Not sure if you need to do it that way or not but this solves the problem: #include <iostream> #include <cmath> using namespace std; int main() { double a, b, c; cout<<“Enter a: “; … Read more
Your main issue is that you declare all your variables as ints, but smp2 and total must hold floating point values. Change your declarations to int smp1; double smp2, total; unsigned int smp3; This way, the types of the variables match up with the conversion specifiers used in the printf and scanf calls. Types matter … Read more
Since you have a low upper bound on n, it’s possible to start off by factoring all of the numbers in the range [1, 3 × 106]. There are many ways to do this reasonably efficiently. One way to do this is to use the Sieve of Eratosthenes or a related sieve to find all … Read more
Python’s math module: A simple Google search would have helped.
Take a look at article https://www.tutorialspoint.com/java/util/java_util_random.htm. All you need to do is to generate pairs of floats in range (-1,1). You should use method nextFloat() from class Random. It will give you numbers in range (0,1). Then multiply it by 2 and subtract 1 and you will have numbers in desired interval.
Look into the python package pandas. Here’s a tutorial: https://pandas.pydata.org/pandas-docs/stable/tutorials.html import pandas as pd list1 = [-50, -40, -30, -20, -10, 0, 1, 2] list2 = [-300, -200, -100, 0, 100, 200, 300, 400] df = pd.DataFrame({‘List 1’: list1, ‘List 2’: list2}) newdf = df.copy() newdf[df > df.median()] = 1 newdf[df < df.median()] = -1 … Read more
Using Howard Hinnant’s free, open-source, header-only date library, it can look like this: #include “date.h” #include “iso_week.h” #include <iostream> int main() { using namespace iso_week::literals; std::cout << date::year_month_day{2017_y/8_w/mon} << ‘\n’; std::cout << date::year_month_day{2017_y/10_w/mon} << ‘\n’; } which outputs: 2017-02-20 2017-03-06 There are also getters for year, month and day on the year_month_day types, and plenty … Read more