hdu 4576 : Robot

Robot

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
Total Submission(s): 1600    Accepted Submission(s): 599

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

 

 

算法：模拟

题目意思：

Robot绕圆(1-n)行走，有m个命令，随机选择方向，第i个命令行走mi个距离。求在l到r的范围内的概率.

思路：

模拟，不超时。

构造两个数组，temp[]记录前一个状态，cnt[]记录当前状态(其初始化为0)，将不为0的temp[]，进行移动，然后加到目的位置的cnt[]中。最后输出ans=temp[l]+...+temp[r];

 

代码：

#include <stdio.h>#include <string.h>#define N 201double cnt[N],temp[N];int main(){    int n,m,l,r,i,j,sum1,w,sum2,pos;    double ans;    while(scanf("%d%d%d%d",&n,&m,&l,&r)!=EOF && (n+m+l+r))    {        memset(cnt,0,sizeof(cnt));        memset(temp,0,sizeof(temp));        temp[1]=1;        for(i=1;i<=m;++i)        {            scanf("%d",&w);//m个w            for(j=1;j<=n;++j)            {                if(temp[j]==0)                    continue;                //顺时针                pos=j+w%n;//确定位置                if(pos>n)                {                    pos-=n;                }                cnt[pos]+=0.5*temp[j];                //逆时针                pos=j-w%n;                if(pos<=0)                {                    pos+=n;                }                cnt[pos]+=0.5*temp[j];            }            for(j=1;j<=n;++j)            {                temp[j]=cnt[j];                cnt[j]=0;            }        }        sum1=sum2=0;        ans=0;        for(i=l;i<=r;++i)        {            ans+=temp[i];        }        printf("%.4lf\n",ans);    }    return 0;}