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Robot

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.

3 1 1 2
1
5 2 4 4
1
2
0 0 0 0

0.5000
0.2500

第一行4个数字表示：一块圆盘上有n个格子，m个操作，l,r表示区间，接下来m行，每行1个数字w，机器人一开始在1号格子，对于每个操作会顺时针或者逆时针移动w格，问你最终停在区间[l,r]的概率。

去年去杭州比赛遇到了这题，当时因为这题超时打铁了，真心感觉自己当时弱爆了！！

现在回过头来看这题，好简单

第i号格子其实就是只能由 i-w号格子 与i+w号格子得来，而且概率各占1半，所以注意边界，用滚动的思想就行了。

#include <iostream>
#include <cstdio>
using namespace std;

const int maxn=210;
double p[maxn],q[maxn];
int n,m,l,r;

void solve(){
    double ans=0;
    for(int i=0;i<=n;i++) q[i]=p[i]=0;
    p[0]=1.0;
    for(int t=0;t<m;t++){
        int w;
        scanf("%d",&w);
        for(int i=0;i<n;i++){
            if(p[i]>0){
                q[(i+w)%n]+=p[i]*0.5;
                q[((i-w)%n+n)%n ]+=p[i]*0.5;
            }

        }
        for(int i=0;i<n;i++){
            p[i]=q[i];
            q[i]=0;
        }
    }
    for(int i=l-1;i<r;i++) ans+=p[i];
    printf("%.4lf\n",ans);
}

int main(){
    while(scanf("%d%d%d%d",&n,&m,&l,&r)!=EOF && (n||m||l||r) ){
        solve();
    }
    return 0;
}

HDU 4576 Robot（概率题）,布布扣,bubuko.com

HDU 4576 Robot（概率题）

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原文地址：http://blog.csdn.net/a1061747415/article/details/36633147