HDU 4576 Robot

Robot

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 2776    Accepted Submission(s): 948

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

 

 解题：dp。。。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
double dp[2][210];
int main() {
    int n,m,l,r,tmp,i,cur;
    double ans = 0.0;
    while(scanf("%d %d %d %d",&n,&m,&l,&r),n||m||l||r){
        dp[0][0] = 1;
        for(int i = 1; i < n; i++) dp[0][i] = 0;
        cur = 0;
        while(m--){
            scanf("%d",&tmp);
            for(i = 0; i < n; i++)
                dp[cur^1][i] = 0.5*dp[cur][(i-tmp+n)%n] + 0.5*dp[cur][(i+tmp)%n];
            cur ^= 1;
        }
        ans = 0;
        for(i = l-1; i < r; i++)
            ans += dp[cur][i];
        printf("%.4f\n",ans);
    }
    return 0;
}