HDU4576-Robot(概率DP+滚动数组)

Michael has a telecontrol robot. One day he put the robot on a loop with n cells. The cells are numbered from 1 to n clockwise.

At first the robot is in cell 1. Then Michael uses a remote control to send m commands to the robot. A command will make the robot walk some distance. Unfortunately the direction part on the remote control is broken, so for every command the robot will chose a direction(clockwise or anticlockwise) randomly with equal possibility, and then walk w cells forward.
Michael wants to know the possibility of the robot stopping in the cell that cell number >= l and <= r after m commands.
Input
There are multiple test cases.
Each test case contains several lines.
The first line contains four integers: above mentioned n(1≤n≤200) ,m(0≤m≤1,000,000),l,r(1≤l≤r≤n).
Then m lines follow, each representing a command. A command is a integer w(1≤w≤100) representing the cell length the robot will walk for this command.
The input end with n=0,m=0,l=0,r=0. You should not process this test case.
Output
For each test case in the input, you should output a line with the expected possibility. Output should be round to 4 digits after decimal points.
Sample Input
3 1 1 2
1
5 2 4 4
1
2
0 0 0 0
Sample Output
0.5000
0.2500

分析：

题意：
第一行输入4个数：n,m.l.r；n表示圆盘的值是从1~n,然后是m次移动，最后l和r分别是区间的左右端点；然后是m行，每行表示这次移动的距离w;输出移动m次最后在l到r这个区间的概率。

解析：
我们用dp[i][i%2]表示移动k次出现在j这个位置的概率；所以我们可以得出状态转移方程：
dp[j][i%2]=(dp[(j-w+n)%n][!(i%2)]+dp[(j+w)%n][!(i%2)])*0.5;

注意：这道题很卡时间，开始没有考虑到（j+w）%n==0的情况，着使得答案错误，然后我用到了memset()这个函数，但是一直超时！！所以，千万不要用这个函数！

代码：

#include<iostream>
#include<cstdio>

using namespace std;

float dp[205][2];

int main()
{
	int n,m,l,r,w;
	while(scanf("%d%d%d%d",&n,&m,&l,&r)&&(n+m+l+r))
	{
		for(int j=0;j<n;j++)
		{
			dp[j][1]=dp[j][0]=0;
		}
		dp[0][0]=1.0;
		for(int i=1;i<=m;i++)
		{
			scanf("%d",&w);
			for(int j=0;j<n;j++)
			{
				dp[j][i%2]=(dp[(j+n-w)%n][!(i%2)]+dp[(j+w)%n][!(i%2)])*0.5;
			}
		}
		float sum=0;
		for(int i=l-1;i<r;i++)
		{
			sum+=dp[i][m%2];
		}
		printf("%.4lf\n",sum);
	}
	return 0;
}