Wunder Fund Round 2016 (Div. 1 + Div. 2 combined) CF618A A. Slime Combining

A. Slime Combining

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

Your friend recently gave you some slimes for your birthday. You have n slimes all initially with value 1.

You are going to play a game with these slimes. Initially, you put a single slime by itself in a row. Then, you will add the other n - 1 slimes one by one. When you add a slime, you place it at the right of all already placed slimes. Then, while the last two slimes in the row have the same value v, you combine them together to create a slime with value v + 1.

You would like to see what the final state of the row is after you’ve added all n slimes. Please print the values of the slimes in the row from left to right.

Input

The first line of the input will contain a single integer, n (1 ≤ n ≤ 100 000).

Output

Output a single line with k integers, where k is the number of slimes in the row after you’ve finished the procedure described in the problem statement. The i-th of these numbers should be the value of the i-th slime from the left.

Sample test(s)

input

1

output

1

input

2

output

2

input

3

output

2 1

input

8

output

4

Note

In the first sample, we only have a single slime with value 1. The final state of the board is just a single slime with value 1.

In the second sample, we perform the following steps:

Initially we place a single slime in a row by itself. Thus, row is initially 1.

Then, we will add another slime. The row is now 1 1. Since two rightmost slimes have the same values, we should replace these slimes with one with value 2. Thus, the final state of the board is 2.

In the third sample, after adding the first two slimes, our row is 2. After adding one more slime, the row becomes 2 1.

In the last sample, the steps look as follows:

1

2

2 1

3

3 1

3 2

3 2 1

4

题意：

看样例的话题目还是比较好理解，大概就是第一个数为1，之后每次操作都在当前数后面的一个数字再加1，如果最右边的两个数的值相等，则最后一个数消除，前一个数在原来值的基础上再加1。现给出数n，求经过n次操作后得到的一串数列是什么。

思路：

方法其实很简单，把前几个数给写出来，如果能看出是二进制规律的话那就so easy啦！二进制从i = 20开始往前遍历，如果pow(2,i * 1.0) <= n，则把i + 1输出，并把n减去pow(2,i * 1.0) ，当n == 0时，结束循环并退出。切记是i + 1！比赛的时候这个坑找了宝宝半天有没有思密达！

#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
while(cin>>n)
{
for(int i = 20; i >= 0; i--)
{
if(pow(2,i * 1.0) <= n)
{
n -= pow(2,i * 1.0);
if(!n)
{
cout<<i + 1<<endl;
break;
}
cout<<i + 1<<" ";
}
}
}
return 0;
}