首页 > 代码库 > HDU 4576 Robot（概率题）

HDU 4576 Robot（概率题）

2024-07-10 16:17:56   226人阅读

Robot


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
 

题目大意：

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。


解题思路：

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。


解题代码：

#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;
}






联系
我们