zoukankan      html  css  js  c++  java
  • LightOJ 1030 数学期望

    Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

    Description

    You are in a cave, a long cave! The cave can be represented by a 1 x N grid. Each cell of the cave can contain any amount of gold.

    Initially you are in position 1. Now each turn you throw a perfect 6 sided dice. If you get X in the dice after throwing, you add X to your position and collect all the gold from the new position. If your new position is outside the cave, then you keep throwing again until you get a suitable result. When you reach the Nth position you stop your journey. Now you are given the information about the cave, you have to find out the expected number of gold you can collect using the given procedure.

    Input

    Input starts with an integer T (≤ 100), denoting the number of test cases.

    Each case contains a blank line and an integer N (1 ≤ N ≤ 100) denoting the dimension of the cave. The next line contains N space separated integers. The ith integer of this line denotes the amount of gold you will get if you come to the ith cell. You may safely assume that all the given integers will be non-negative and no integer will be greater than 1000.

    Output

    For each case, print the case number and the expected number of gold you will collect. Errors less than 10-6 will be ignored.

    Sample Input

    3

    1

    101

    2

    10 3

    3

    3 6 9

    Sample Output

    Case 1: 101.0000000000

    Case 2: 13.000

    Case 3: 15

    题意:走到某个点时就掷一个6面骰子,掷到几就在向前走几步,走到的点宝藏就能拿走,最后要求数学期望

    做法:那么这个点的期望dp[i] = dp[i  +1] /6  + dp[i + 2] / 6 + dp[i + 3] /6 + dp[i + 4] / 6 + dp[i + 5] / 6 + dp[i + 6] / 6 + dp[i];

    对于小于6的点另做处理

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<cmath>
    #include<cctype>
    #include<queue>
    #include<stack>
    #include<vector>
    #include<algorithm>
    
    using namespace std;
    typedef long long LL;
    #define N 210
    #define INF 0x3f3f3f3f
    
    double a[N], dp[N];
    
    int main()
    {
        int T, n, cas=1;
        scanf("%d", &T);
    
        while(T--)
        {
            scanf("%d", &n);
            memset(dp, 0, sizeof(dp));
            for(int i=0; i<n; i++)
                scanf("%lf", &a[i]);
    
            dp[n-1]=a[n-1];
            for(int i=n-2; i>=0; i--)
            {
                dp[i]=a[i];
                int t=6;
                if(n-1-i<6)
                    t=n-1-i;
                for(int j=1; j<=t; j++)
                    dp[i]+=dp[i+j]/t;
            }
            printf("Case %d: %.10f
    ", cas++, dp[0]);
        }
        return 0 ;
    }
  • 相关阅读:
    VS2010中经常使用的快捷键
    IE无法打开internet网站已终止操作的解决的方法
    Spring3.0 AOP 具体解释
    Java 反射机制[Method反射]
    软件測试自学指南---从入门到精通
    Java中Map的使用
    tracert路由跟踪命令分析判断
    C++ Primer 学习笔记_32_STL实践与分析(6) --再谈string类型(下)
    ORACLE触发器具体解释
    Androidclient推断server是否开启 HttpHostException解决方式
  • 原文地址:https://www.cnblogs.com/9968jie/p/5789704.html
Copyright © 2011-2022 走看看