• HDU 4576 Robot (很水的概率题)


    Robot

    Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)
    Total Submission(s): 158    Accepted Submission(s): 46


    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
     
     
     
     
     
    按照概率DP过去,比较暴力,T_T
     
     1 /* **********************************************
     2 Author      : kuangbin
     3 Created Time: 2013/8/10 11:51:05
     4 File Name   : F:2013ACM练习比赛练习2013杭州邀请赛重现1001.cpp
     5 *********************************************** */
     6 
     7 #include <stdio.h>
     8 #include <string.h>
     9 #include <iostream>
    10 #include <algorithm>
    11 #include <vector>
    12 #include <queue>
    13 #include <set>
    14 #include <map>
    15 #include <string>
    16 #include <math.h>
    17 #include <stdlib.h>
    18 using namespace std;
    19 double dp[2][220];
    20 int main()
    21 {
    22     //freopen("in.txt","r",stdin);
    23     //freopen("out.txt","w",stdout);
    24     int n,m,l,r;
    25     while(scanf("%d%d%d%d",&n,&m,&l,&r) == 4)
    26     {
    27         if(n == 0 && m == 0 && l == 0 && r == 0)break;
    28         dp[0][0] = 1;
    29         for(int i = 1;i < n;i++)dp[0][i] = 0;
    30         int now = 0;
    31         while(m--)
    32         {
    33             int v;
    34             scanf("%d",&v);
    35             for(int i = 0;i < n;i++)
    36                 dp[now^1][i] = 0.5*dp[now][(i-v+n)%n] + 0.5*dp[now][(i+v)%n];
    37             now ^= 1;
    38         }
    39         double ans = 0;
    40         for(int i = l-1;i < r;i++)
    41             ans += dp[now][i];
    42         printf("%.4lf
    ",ans);
    43     }
    44     return 0;
    45 }
     
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  • 原文地址:https://www.cnblogs.com/kuangbin/p/3250367.html
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