首页 > 其他 > 详细

codeforces Gravity Flip 题解

时间:2018-02-04 12:02:22      阅读:193      评论:0      收藏:0      [点我收藏+]

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.

技术分享图片

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 
非常有趣。非常有新意的题目。

原来能够这么改变gravity,形成这种效果的。有意思。

只是题解却是非常容易的,就是一个排序问题,O(∩_∩)O~,可是没看到这种题目还真没想过。

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;
}




codeforces Gravity Flip 题解

原文:https://www.cnblogs.com/llguanli/p/8412560.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!