HDU 4576 Robot（概率dp）

Robot

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 5906    Accepted Submission(s): 1754

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

分析

code

 #include<cstdio>
 #include<algorithm>
 #include<cstring>

 using namespace std;

 double f[][];
 // f[i][j]到第i次操作，位置j上的概率 

 int main() {

     int n,m,l,r;
     while (~scanf("%d%d%d%d",&n,&m,&l,&r) && n+m+l+r) {
         memset(f,,sizeof(f));
         f[][] = 1.0;
         int cur = ;
         for (int w,i=; i<=m; ++i) {
             scanf("%d",&w);
             w = w%n;
             cur = cur^;
             for (int k=; k<=n; ++k) {
                 f[cur][k] = f[cur^][k+w>n?k+w-n:k+w]/2.0 + f[cur^][k-w<?k-w+n:k-w]/2.0;
             }
         }
         double ans = 0.0;
         for (int i=l; i<=r; ++i) ans += f[cur][i];
         printf("%.4lf\n",ans);
     }
     return ;
 }