codeforces Gravity Flip 题解

最新推荐文章于 2020-03-14 19:33:49 发布

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Little Chris 在物理课上感到无聊，于是制作了一个玩具盒子来保持自己忙碌。这个盒子有一个特殊的能力，能够改变重力方向。盒子中有 n 列玩具立方体，每列的立方体数量各不相同。当 Chris 将重力切换到右侧时，立方体会重新排列位置。此题要求根据初始状态，找出重力切换后的立方体分布。

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Little Chris is bored during his physics lessons (too easy), so he has built a toy box to keep himself occupied. The box is special, since it has the ability to change gravity.

There are n columns of toy cubes in the box arranged in a line. The i-th column contains ai cubes. At first, the gravity in the box is pulling the cubes downwards. When Chris switches the gravity, it begins to pull all the cubes to the right side of the box. The figure shows the initial and final configurations of the cubes in the box: the cubes that have changed their position are highlighted with orange.

$\text{[math]}$

Given the initial configuration of the toy cubes in the box, find the amounts of cubes in each of the n columns after the gravity switch!

Input

The first line of input contains an integer n (1 ≤ n ≤ 100), the number of the columns in the box. The next line contains n space-separated integer numbers. The i-th number ai (1 ≤ ai ≤ 100) denotes the number of cubes in the i-th column.

Output

Output n integer numbers separated by spaces, where the i-th number is the amount of cubes in the i-th column after the gravity switch.

Sample test(s)

input
4
3 2 1 2

output
1 2 2 3 

很有趣，很有新意的题目。

原来可以这么改变gravity，形成这样的效果的，有意思。

不过题解却是很容易的，就是一个排序问题，O(∩_∩)O~，但是没看到这样的题目还真没想过。

void GravityFlip()
{
	unsigned n;
	cin>>n;
	int *A = new int[n];

	for (unsigned i = 0; i < n; i++)
	{
		cin>>A[i];
	}

	sort(A, A+n);
	for (unsigned i = 0; i < n; i++)
	{
		cout<<A[i]<<' ';
	}

	delete [] A;
}