Uva 11384 - 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.

 

SAMPLE INPUT	OUTPUT FOR SAMPLE INPUT
1 2 3	1 2 2

 

Problemsetter: Md. Mahbubul Hasan

很水的一道题，但关键的是刚好用到了老师今天讲的分治法，让我对分治的概念有了新的认识。

题目的意思是给定一个N，你每次可以再屏幕上选取一个或几个数，让他们同时减去任意一个数，输出使所有数字都变成0的最小次数。

试几个样例就会有所发现：

N == 7时，有1 2 3 4 5 6 7

选取4 5 6 7，使它们同时减去4，得到1 2 3 0 1 2 3

处理这几个数，等价于处理1 2 3

同理处理1 2 3即可

代码如下：

#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#define MAXN 100
#define ll long long
using namespace std;
ll times = 0;
int method(ll n){
	if(n == 1)
	 return(++times);
		times++;
		return method(n/2);
}

int main(void){
     ll n;
	 while(cin >> n){
	 	times = 0;
	 	cout << method(n) << endl;
	 } 
	 return 0;
}