integer-arithmetic

UVA Problem no. 10055, Hashmat the Brave Warrior , probably the easiest problem there. The input consists of a series of pairs of unsigned integers ≤ 2^32 (thus mandating the use of 64bit integers…) For each pair the task is to print out the difference between the greater and the lesser integer. According to the statistics , the fastest solutions run in below 0.01 sec. However, all my attempts to solve this typically run in 0.02 sec, with probably random deviations of ± 0.01 sec. I tried: #include <cstdint> #include <iostream> using namespace std; int main() { ios_base::sync_with_stdio(false);

I need to traverse all pairs i,j with 0 <= i < n , 0 <= j < n and i < j for some positive integer n . Problem is that I can only loop through another variable, say k . I can control the bounds of k . So the problem is to determine two arithmetic methods, f(k) and g(k) such that i=f(k) and j=g(k) traverse all admissible pairs as k traverses its consecutive values. How can I do this in a simple way? I think I got it (in Python): def get_ij(n, k): j = k // (n - 1) # // is integer (truncating) division i = k - j * (n - 1) if i >= j: i = (n - 2) - i j = (n - 1) - j return i, j for n in range(2, 6):

Closed . This question needs to be more focused. It is not currently accepting answers. Learn more . Want to improve this question? Update the question so it focuses on one problem only by editing this post . From the comments on my answer here , the question was asked (paraphrase): Write a Python program to find a 4 digit whole number, that when multiplied to itself, you get an 8 digit whole number who's last 4 digits are equal to the original number. I will post my answer, but am interested in a more elegant solutions concise but easily readable solution! (Would someone new-ish to python be

The following Python 3.x integer multiplication takes on average between 1.66s and 1.77s: import time start_time = time.time() num = 0 for x in range(0, 10000000): # num += 2 * (x * x) num += 2 * x * x print("--- %s seconds ---" % (time.time() - start_time)) if I replace 2 * x * x with 2 *(x * x) , it takes between 2.04 and 2.25 . How come? On the other hand it is the opposite in Java: 2 * (x * x) is faster in Java. Java test link: Why is 2 * (i * i) faster than 2 * i * i in Java? I ran each version of the program 10 times, here are the results. 2 * x * x | 2 * (x * x) ------------------------