HDOJ 4576

Robot

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1216    Accepted Submission(s): 470

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

 

直接去模拟即可，滚动数组。

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>


using namespace std;
//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1|1
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
#define mid ((l + r) >> 1)
#define mk make_pair
const int MAXN = 100000 + 100;
const int maxw = 10000000 + 20;
const int MAXNNODE = 10000 +10;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
#define mod 10007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pii;
D f[2][210];
int main()
{
    //ios::sync_with_stdio(false);
#ifdef Online_Judge
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
#endif // Online_Judge
    int n , m  , l , r , w;
    while(~scanf("%d%d%d%d" , &n , &m , &l , &r) , n)
    {
        int now = 0;
        clr(f[0] , 0);
        f[0][1] = 1;
        FORR(k , 1 , m)
        {
            scanf("%d", &w);
            now ^= 1;
            FORR(i , 1 , n)///逆时针为减w，顺时针为加w
            {
                f[now][i] = (f[now ^ 1][i - w + (i - w >= 1 ? 0 : n)] + f[now ^ 1][i + w - (i + w <= n ? 0 : n)]) * 0.5;
            }

        }
        D ans = 0;
            FORR(i , l , r)ans += f[now][i];
            FORR(i , 0 , n)printf("  %lf %lf\n",f[0][i] , f[1][i]);
            printf("%.4lf\n" , ans);
    }
    return 0;
}