CodeForces - 405B（Domino Effect）

最新推荐文章于 2019-04-03 19:53:41 发布

原创最新推荐文章于 2019-04-03 19:53:41 发布 · 296 阅读

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#模拟 #CodeForces

本文解析了一道名为DominoEffect的编程题，通过模拟过程找出最终保持直立状态的多米诺骨牌数量。文章提供了完整的代码实现，并对关键步骤进行了说明。

Little Chris knows there's no fun in playing dominoes, he thinks it's too random and doesn't require skill. Instead, he decided to play with the dominoes and make a "domino show".

Chris arranges n dominoes in a line, placing each piece vertically upright. In the beginning, he simultaneously pushes some of the dominoes either to the left or to the right. However, somewhere between every two dominoes pushed in the same direction there is at least one domino pushed in the opposite direction.

After each second, each domino that is falling to the left pushes the adjacent domino on the left. Similarly, the dominoes falling to the right push their adjacent dominoes standing on the right. When a vertical domino has dominoes falling on it from both sides, it stays still due to the balance of the forces. The figure shows one possible example of the process.

$\text{[math]}$

Given the initial directions Chris has pushed the dominoes, find the number of the dominoes left standing vertically at the end of the process!

Input

The first line contains a single integer n (1 ≤ n ≤ 3000), the number of the dominoes in the line. The next line contains a character string s of length n. The i-th character of the string si is equal to

"L", if the i-th domino has been pushed to the left;
"R", if the i-th domino has been pushed to the right;
".", if the i-th domino has not been pushed.

It is guaranteed that if si = sj = "L" and i < j, then there exists such k that i < k < j and sk = "R"; if si = sj = "R" and i < j, then there exists such k that i < k < j and sk = "L".

Output

Output a single integer, the number of the dominoes that remain vertical at the end of the process.

   Examples
 
     input
   
14
.L.R...LR..L..

     output
   
4

     input
   
5
R....

     output
   
0

     input
   
1
.

     output
   
1

Note

The first example case is shown on the figure. The four pieces that remain standing vertically are highlighted with orange.

In the second example case, all pieces fall down since the first piece topples all the other pieces.

In the last example case, a single piece has not been pushed in either direction.

题目连接：点击打开链接

题意：就是一个卡诺牌，给你个顺序，然后判断下有几个竖着的，一个模拟题，L代表往左倒，R代表往右倒，‘.’就代表普通的，直接模拟走就行了；

#include<stdio.h>
#include<iostream> 
#include <algorithm>
#include<string.h>
#include<math.h>
#include<queue>
#include<set>
#define LL long long
#define INF 0x3f3f3f3f
#define s(n) scanf("%d",&n)
using namespace std; 
int main()
{
	int n,i;
	char st[3100];
	while(scanf("%d",&n)!=EOF)
	{
		scanf("%s",&st);
		int ans1=0;//记录到L或者R之间有多少个.
		int ans=0;//记录总共有多少个可以竖着 
		int flag=0;//碰到一个往右的就改成1 
		for(int i=0;i<n;i++)
		{
			if(st[i]=='.')
				ans1++;
			if(st[i]=='L')//往左边歪倒  所以过来这一路的所有的.都清空 
			{
				if(flag)//说明右边有一个配对的 
				{
					if(ans1%2)//中间的做柱子 
						ans=ans+1;
				} 
				ans1=0;//判断完了
				flag=0;
			} 
			if(st[i]=='R')//开始储存当前的点 
			{
				ans=ans+ans1;
				ans1=0;
				flag=1;
			}
		}
		if(flag)//最后剩下一个往右的 所有往右记录的.都清空 
			ans1=0;
		printf("%d\n",ans+ans1);
	}
	return 0;
}

模拟题目就是需要一个思维，如果卡思维就完犊子了~