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[2][220];
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) == 4)
    {
        if(n == 0 && m == 0 && l == 0 && r == 0)break;
        dp[0][0] = 1;
        for(int i = 1;i < n;i++)dp[0][i] = 0;
        int now = 0;
        while(m--)
        {
            int v;
            scanf("%d",&v);
            for(int i = 0;i < n;i++)
                dp[now^1][i] = 0.5*dp[now][(i-v+n)%n] + 0.5*dp[now][(i+v)%n];
            now ^= 1;
        }
        double ans = 0;
        for(int i = l-1;i < r;i++)
            ans += dp[now][i];
        printf("%.4lf
",ans);
    }
    return 0;
}