HDU 4576 Robot 概率DP 水题

　　　　　　　　　　　　　　　　　　 　　　Robot

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 3851    Accepted Submission(s): 1246

Problem Description

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

 

Source

2013ACM-ICPC杭州赛区全国邀请赛

 

题意：给出一个环，上面分成n个区域，分别编号为1~n，有一个指针，刚开始的时候指向1的位置

现在有m个操作，每一个操作输入一个w，指针转动w个单位，顺时针转，和逆时针转的概率都是0.5，

现在问m个操作后，指针指向区间[l,r]的概率是多少

dp[i][j]表示第i个命令后，指针指向区域j的概率

这里命令数m很大，而我们递推的时候只需要保存前面一个的状态，所以i%2，节省空间。

#include<cstdio>
#include<cstring>

using namespace std;

double dp[2][210];

int main()
{
    int n,m,l,r;
    while(scanf("%d%d%d%d",&n,&m,&l,&r))
    {
        if(!n&&!m&&!l&&!r)
            break;
        memset(dp,0,sizeof dp);
        dp[0][1]=1.0;
        int w;
        int a,b;
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&w);
            for(int j=1;j<=n;j++)
            {
                a=j-w;
                if(a<1)
                    a+=n;
                b=j+w;
                if(b>n)
                    b-=n;
                dp[i%2][j]=dp[(i-1)%2][a]*0.5+dp[(i-1)%2][b]*0.5;
            }
        }
        double ans=0.0;
        int mm=m%2;
        for(int i=l;i<=r;i++)
            ans+=dp[mm][i];

        printf("%.4f
",ans);
    }
    return 0;
}