UVa11384 - 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

题目大意：给定一个序列1,2,3,...n,每次你可以选取一个或者多个数减去同一个数（必须是正数），用尽量少的步数使得序列所有值变为0。

题解：对于序列1~n,我们对于序列1~n/2不进行操作,对于后半部分,即n/2+1~n全部减去n/2+1，我们得到新的序列1,2,3...n/2,0,1,2...(n-1)/2。也就等价于1,2,3..n/2,因为后半部分序列的每一个值在前半部分序列中都存在。所以我们可以得出一个公式:f(n)=f(n/2)+1，边界条件是n=1。

View Code

 1 #include<stdio.h>

 2 long f(long n)

 3 {

 4     return n==1?1:f(n/2)+1;

 5 }

 6 int main(void)

 7 {

 8     long n;

 9     while(scanf("%ld",&n)!=EOF)

10     printf("%ld\n",f(n));

11     return 0;

12 }