Consider the following problem statement:
There are a 100 monkeys, each has a key that can open or close 100 doors. The monkeys come in sequence from 1 to 100 and open or close doors with numbers from their respective sequence number's multiplication table. i.e. monkey #2 can open all even numbered doors while monkey #50 can open only the doors numbered 50 and 100.
What are the doors left opened after all monkeys have done their monkey business?
Solution: Idea: Door i will be open if the number of factors to i are odd, else it will be closed. Since there are only 100 doors to check and we have to check all doors, a nested loop will be a very effective solution:
let mut doors = [false; 100];
for i in 1..=100 { // iterate through monkeys
for j in 1..=100/i { // iterate through multiplication table for each monkey, numbers <100
doors[(i*j) - 1] = !doors[(i*j) - 1]; // arrays are 0 indexed, so subtract 1
}
}
println!("{:?}", doors);