HDU 4576 Robot （很水的概率题）

Robot

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 158    Accepted Submission(s): 46

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

 

 

 

 

 

按照概率DP过去，比较暴力，T_T

 

 1 /* **********************************************

 2 Author      : kuangbin

 3 Created Time: 2013/8/10 11:51:05

 4 File Name   : F:\2013ACM练习\比赛练习\2013杭州邀请赛重现\1001.cpp

 5 *********************************************** */

 6 

 7 #include <stdio.h>

 8 #include <string.h>

 9 #include <iostream>

10 #include <algorithm>

11 #include <vector>

12 #include <queue>

13 #include <set>

14 #include <map>

15 #include <string>

16 #include <math.h>

17 #include <stdlib.h>

18 using namespace std;

19 double dp[2][220];

20 int main()

21 {

22     //freopen("in.txt","r",stdin);

23     //freopen("out.txt","w",stdout);

24     int n,m,l,r;

25     while(scanf("%d%d%d%d",&n,&m,&l,&r) == 4)

26     {

27         if(n == 0 && m == 0 && l == 0 && r == 0)break;

28         dp[0][0] = 1;

29         for(int i = 1;i < n;i++)dp[0][i] = 0;

30         int now = 0;

31         while(m--)

32         {

33             int v;

34             scanf("%d",&v);

35             for(int i = 0;i < n;i++)

36                 dp[now^1][i] = 0.5*dp[now][(i-v+n)%n] + 0.5*dp[now][(i+v)%n];

37             now ^= 1;

38         }

39         double ans = 0;

40         for(int i = l-1;i < r;i++)

41             ans += dp[now][i];

42         printf("%.4lf\n",ans);

43     }

44     return 0;

45 }