When you first made the computer to print the sentence “Hello World!”, you felt so happy, not
knowing how complex and interesting the world of programming and algorithm will turn out to be.
Then you did not know anything about loops, so to print 7 lines of “Hello World!”, you just had to
copy and paste some lines. If you were intelligent enough, you could make a code that prints “Hello
World!” 7 times, using just 3 paste commands. Note that we are not interested about the number
of copy commands required. A simple program that prints “Hello World!” is shown in Figure 1. By
copying the single print statement and pasting it we get a program that prints two “Hello World!”
lines. Then copying these two print statements and pasting them, we get a program that prints four
“Hello World!” lines. Then copying three of these four statements and pasting them we can get a
program that prints seven “Hello World!” lines (Figure 4). So three pastes commands are needed in
total and Of course you are not allowed to delete any line after pasting. Given the number of “Hello
World!” lines you need to print, you will have to find out the minimum number of pastes required to
make that program from the origin program shown in Figure 1.

臥槽同志們這題真的不是寬搜嗎？！

我一眼看上去就覺得這是寬搜模板啊，然後就寬搜了，然後就ac了。結果被隊友說思路清奇。好的我知道你們都是O(1)過的。網上的也都是貪心或者找規律………..
我以為全世界寫這題都會用寬搜，結果只有我是奇葩嗎orz

ac程式碼:

public class Main {
    static int[] step = new int[10005];
    static Set<Integer> visited = new HashSet<Integer>();

    public static void main(String[] args) {
        Scanner reader = new
 
 Scanner(System.in);
        int n = reader.nextInt();
        int cases = 1;
        slove(10001);
        while (n > 0) {
            System.out.print("Case " + (cases++) + ": ");
            System.out.println(step[n]);
            n = reader.nextInt();
        }
    }

    public static void slove
 
(int n) {
        Queue<Integer> wait = new LinkedList<Integer>();
        wait.add(1);
        visited.add(1);
        int cur = 0;
        int next = 0;
        while (!wait.isEmpty()) {
            cur = wait.poll();
            for (int i = 1; i <= cur; i++) {
                next = cur + i;
                if (next >= 0 && next <= n) {//其實可以不用判範圍....
                    if (visited.add(next)) {
                        step[next] = step[cur] + 1;
                        wait.add(next);
                    }
                    if (next == n) {
                        return;
                    }
                }
            }
        }
    }
}

這題真的跟farmer john找牛不是一個套路嗎？！