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

 

 /* **********************************************
 Author      : kuangbin
 Created Time: 2013/8/10 11:51:05
 File Name   : F:\2013ACM练习\比赛练习\2013杭州邀请赛重现\1001.cpp
 *********************************************** */

 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 using namespace std;
 double dp[][];
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int n,m,l,r;
     while(scanf("%d%d%d%d",&n,&m,&l,&r) == )
     {
         if(n ==  && m ==  && l ==  && r == )break;
         dp[][] = ;
         for(int i = ;i < n;i++)dp[][i] = ;
         int now = ;
         while(m--)
         {
             int v;
             scanf("%d",&v);
             for(int i = ;i < n;i++)
                 dp[now^][i] = 0.5*dp[now][(i-v+n)%n] + 0.5*dp[now][(i+v)%n];
             now ^= ;
         }
         double ans = ;
         for(int i = l-;i < r;i++)
             ans += dp[now][i];
         printf("%.4lf\n",ans);
     }
     return ;
 }