UVA11384正整数序列Help is needed for Dexter

Problem H

Help is needed for Dexter

Time Limit: 3 Second

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 and Output:

Input consists of several lines each with N such that 1 ≤ N ≤ 1,000,000,000. Input will be terminated by end of file. For each N output L in separate lines.
中文大意：
给定正整数n，你的任务是用最少的操作次数把序列1,2,……,n中所有数都变成0。每次操作可从序列中选择一个或多个整数，同时减去一个相同的正整数。比如，1,2,3 可以把2和3同时减小2，得到1,0,1。

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思路： 我们只考虑操作一次的时候，操作一次最大的效果就是剪掉一半，
比如1 2 3 4 5 6 7 变成1 2 3 0 1 2 3 等价于1 2 3直接减少一半（0不影响，因为最后变值为0的时候，0不管是加一个数或减一个数都是不影响的），那么在把减少后等等价状态1 2 3在作同样的操作直到只剩一个的时候就ok了。

#include <cstdio>
using namespace std;

int f(int n){
	return n == 1 ? 1:f(n/2)+1;
	 
} 

int main(){
	int n;
	while(scanf("%d",&n) == 1)
	printf("%d\n",f(n));
	return 0;
}