# 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

 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 }

posted on 2013-08-10 18:04 kuangbin 阅读(...) 评论(...) 编辑 收藏

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