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[2][210];
// 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,0,sizeof(f));
        f[0][1] = 1.0;
        int cur = 0;        
        for (int w,i=1; i<=m; ++i) {
            scanf("%d",&w);
            w = w%n;
            cur = cur^1;
            for (int k=1; k<=n; ++k) {
                f[cur][k] = f[cur^1][k+w>n?k+w-n:k+w]/2.0 + f[cur^1][k-w<1?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 0;    
}