Imagine you have a special keyboard with the following keys:
Key 1: (A): Print one 'A' on screen.
Key 2: (Ctrl-A): Select the whole screen.
Key 3: (Ctrl-C): Copy selection to buffer.
Key 4: (Ctrl-V): Print buffer on screen appending it after what has already been printed.
Now, you can only press the keyboard for N times (with the above four keys), find out the maximum numbers of 'A' you can print on screen.
Example 1:
Input: N = 3 Output: 3 Explanation: We can at most get 3 A's on screen by pressing following key sequence: A, A, A
Example 2:
Input: N = 7 Output: 9 Explanation: We can at most get 9 A's on screen by pressing following key sequence: A, A, A, Ctrl A, Ctrl C, Ctrl V, Ctrl V
Note:
struct Solution;
impl Solution {
fn max_a(n: i32) -> i32 {
let n = n as usize;
let mut dp = vec![0; n + 1];
for i in 0..=n {
dp[i] = i as i32;
if i > 2 {
for j in 1..i - 2 {
dp[i] = dp[i].max(dp[j] * (i - j - 1) as i32);
}
}
}
dp[n]
}
}
#[test]
fn test() {
let n = 3;
let res = 3;
assert_eq!(Solution::max_a(n), res);
let n = 7;
let res = 9;
assert_eq!(Solution::max_a(n), res);
}