package Classify.DP.Medium;
import org.junit.jupiter.api.Test;
/**
-
Initially on a notepad only one character 'A' is present. You can perform two operations on this notepad for each step:
Copy All: You can copy all the characters present on the notepad (partial copy is not allowed).
Paste: You can paste the characters which are copied last time.
Given a number n. You have to get exactly n 'A' on the notepad by performing the minimum number of steps permitted. Output the minimum number of steps to get n 'A'.Example 1:
Input: 3
Output: 3
Explanation:
Intitally, we have one character 'A'.
In step 1, we use Copy All operation.
In step 2, we use Paste operation to get 'AA'.
In step 3, we use Paste operation to get 'AAA'.
Note:
The n will be in the range [1, 1000].
*/
public class KeysKeyboard {
/**
* 這題的思路就是 dp 代表的就是當前的 A 的數量,然后我們優先考慮上一步復制過的情況，因為這樣能夠最快的填滿 n 個 A ，但是使用這種情況要注意的就是
* 我們可能上部賦值我們最后收不了尾，所以必須有一個判斷就是我們能用上一步的復制的內容填滿剩下的 A ，也就是 n % dp[i - 1] == 0
* 里面的內容的意思就是我們使用了上一步的復制的內容那么我們的字符串就直接翻倍，而且我們進行了一次復制，那么我們 count 就得加2
* @param n
* @return
*/
public int minSteps(int n) {
if (n == 1) {
return 0;
}
int[] dp = new int[1000];
dp[0] = 1;
int paste = 1;
int i = 0;
int count = 1;
while (dp[i] <= n) {
++i;
if (n % dp[i - 1] == 0) {
paste = dp[i - 1];
dp[i] = dp[i - 1] * 2;
++count;
if (paste != 1) {
++count;
}
} else {
dp[i] = dp[i - 1] + paste;
++count;
}
if (dp[i] == n) {
return count;
}
}
return count;
}
@Test
public void fun() {
System.out.println(minSteps(10));
}
}
