UVA11384 Help is needed for Dexter (递归、找规律)

Dexter is tired of Dee Dee. So he decided to keep Dee Dee busy in a game. The game he planned for
her is quite easy to play but not easy to win at least not for Dee Dee. But Dexter does not have time
to spend on this silly task, so he wants your help.
There will be a button, when it will be pushed a random number
N
will be chosen by computer.
Then on screen there will be numbers from 1 to
N
. Dee Dee can choose any number of numbers from
the numbers on the screen, and then she will command computer to subtract a positive number chosen
by her (not necessarily on screen) from the selected numbers. Her objective will be to make all the
numbers 0.
For example if
N
= 3, then on screen there will be 3 numbers on screen: 1, 2, 3. Say she now selects
1 and 2. Commands to subtract 1, then the numbers on the screen will be: 0, 1, 3. Then she selects 1
and 3 and commands to subtract 1. Now the numbers are 0, 0, 2. Now she subtracts 2 from 2 and all
the numbers become 0.
Dexter is not so dumb to understand that this can be done very easily, so to make a twist he will
give a limit
L
for each
N
and surely
L
will be as minimum as possible so that it is still possible to win
within
L
moves. But Dexter does not have time to think how to determine
L
for each
N
, so he asks
you to write a code which will take
N
as input and give
L
as output.
Input
Input consists of several lines each with
N
such that 1

N

1
;
000
;
000
;
000. Input will be terminated
by end of le.
Output
For each
N
output
L
in separate lines.
SampleInput
1
2
3
SampleOutput
1
2
2

这道题自己推到了1–9的情况，于是发现就是个二分找。。。于是递归就行了。
一开始我还用dp记忆化一下，发现自己真的年轻了。。

#include<cstdio>int dfs(int n){    if(n==1)return 1;    return dfs(n>>1)+1;}int main(){    int n;    while(~scanf("%d",&n)){        printf("%d\n",dfs(n));    }    return 0;}