Monday, February 21, 2011
Project Euler 1
I'd like to get through as many of the Project Euler problems as I can in the evenings. I won't have a lot of time to dig through the next chapter of the GTK+ book until mid-March I figure and these problems can usually be knocked out fast once you figure out what has to be done.
Problem 1: http://projecteuler.net/index.php?section=problems&id=1
As long as you know how to test for remainders in a division you're golden. No real trick here.
Subscribe to:
Post Comments (Atom)
Fun(?) with number sequences:
ReplyDelete( 1 + 2 + 3 + ... + X )
= X(X+1)/2
So,
( 3 + 6 + ... + L ) +
( 5 + 10 + ... + M ) -
( 15 + 30 + ... + N )
= 3 * ( 1 + 2 + ... + L ) +
5 * ( 1 + 2 + ... + M ) -
15 * ( 1 + 2 + ... + N )
= 3 * L(L+1)/2 + 5 * M(M+1)/2 - 15 * N(N+1)/2
For L=999/3, M=999/5, N=999/15:
// Sum all multiples of 3 or 5 below 1000
// And, make sure not to count multiples of BOTH 3 and 5 twice!
unsigned int limit = 1000;
unsigned int L = (limit-1)/3;
unsigned int M = (limit-1)/5;
unsigned int N = (limit-1)/15;
unsigned int total = (3*(L*(L+1))+5*(M*(M+1))-15*(N*(N+1)))/2;