UVA 11384 Help is needed for Dexter（问题转化 递归）

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.

SAMPLE INPUT

1

2

3

OUTPUT FOR SAMPLE INPUT

1

2

2

题目大意：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中的所有数都变成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。

分析：正整数序列。拿到这个题目最好的方式是自己试一试。当n=6的时候留下1,2,3，而把4,5,6同时减去min{4,5,6}得到序列1,2,3,0,1,2，它等价于1,2,3。换句话说，f(6)=f(3)+1。

　　一般的，为了平衡，我们保留1~n/2，把剩下的数同时减去n/2+1，得到序列，1,2,...,n/2,0,1,...,(n-1)/2，它等价于1,2,...,n/2.因此f(n)=f(n/2)+1.边界是f(1)=1。

代码如下：

#include<cstdio>
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;
}