hdu—4576

Robot

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 5069    Accepted Submission(s): 1494

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 1 215 2 4 4120 0 0 0

 

Sample Output

0.50000.2500

 

Source

2013ACM-ICPC杭州赛区全国邀请赛

 

【分析】

题意：有一个机器人，每个命令是走m步，可以向前可以向后，问最后停在[l,r]之间的概率

数据挺小的...模拟或者概率dp都可以..我这里直接写了一个简单dp...因为数据挺小的直接去dp模拟状态了

对于f[i][j]，i表示状态，j表示点。直接模拟就可以了

【代码】

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std; int main(){    int n,m,l,r;    double f[2][210];    while(~scanf("%d%d%d%d",&n,&m,&l,&r))    {        if(n==0&&m==0&&l==0&&r==0)break;        memset(f,0,sizeof(f));        f[0][1]=1;        int temp1,temp2,temp;    temp=0;        for(int i=1;i<=m;i++)        {            int w;scanf("%d",&w);            for(int j=1;j<=n;j++)            {                temp1=j+w;if (temp1>n) temp1-=n;                temp2=j-w;if (temp2<1) temp2+=n;                f[(temp+1)%2][temp1]+=f[temp][j]*0.5;                f[(temp+1)%2][temp2]+=f[temp][j]*0.5;                f[temp][j]=0;               }            temp=(temp+1)%2;        }        double ans=0;        for(int i=l;i<=r;i++) ans+=f[temp][i];        printf("%.4f\n",ans);    }    return 0;}