【CodeForces 618A】Slime Combining（水题）

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本文探讨了一种独特的游戏机制，玩家通过合成相同数值的元素以增加其价值，从初始的单一元素逐步构建复合状态，直至所有元素被整合。详细解释了实现方法，包括使用栈数据结构来简化元素的合成过程，并提供了代码示例以直观展示整个流程。

Description

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 Input

Input

Output

Input

Output

Input

Output

2 1

Input

Output

题目大意

给你一个整数n表示有n个1，每两个相同的数a可以合成a+1，比如n=3,表示有3个1，其中两个1可以合成2，剩下一个1，所以输出2 1。

思路

实现方法很多，我是用栈来实现的，每次都将当前准备进栈的数同栈顶元素比较，相等则栈顶元素出栈并+1，再重复上述步骤知道当前进栈元素和栈顶元素不相同时，进栈。最后用数组存栈中的数据，逆序输出就好了（因为栈是先进后出）

代码

#include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <stack>

 using namespace std;

 const int maxn=1e5+5;

 int ans[maxn];

 int main()
 {
    int n;
    while(~scanf("%d",&n))
    {
        stack<int> q;
        int temp=-1,tem=-1;
        for(int i=1;i<=n;i++)
        {
            if(!q.empty()) temp=q.top();
            else temp==-1;
            q.push(1);
            while(q.top()==temp)
            {
                tem=q.top()+1;
                q.pop();q.pop();
                if(!q.empty()) temp=q.top();
                else temp==-1;
                q.push(tem);
            }
        }
        int k=q.size();
        for(int i=0;i<k;i++)
        {
            ans[i]=q.top();
            q.pop();
        }
        for(int i=k-1;i>=0;i--)
        {
            if(i==k-1) printf("%d",ans[i]);
            else printf(" %d",ans[i]);
        }
        printf("\n");
    }
    return 0;
 }