lightoj 1030 概率dp

超、凢脫俗 2022-07-28 04:10 159阅读 0赞

1030 - Discovering Gold

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   <td>Time Limit: <span style="color:#B45F04">2 second(s)</span></td> 
   <td>Memory Limit: <span style="color:#B45F04">32 MB</span></td> 
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You are in a cave, a long cave! The cave can be representedby 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 youthrow a perfect **6** sided dice. If you get **X** in the dice afterthrowing, you add **X** to your position and collect all the gold from thenew position. If your new position is outside the cave, then you keep throwingagain until you get a suitable result. When you reach the **Nth**position you stop your journey. Now you are given the information about thecave, you have to find out the **expected** number of gold you can collectusing 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 linecontains **N** space separated integers. The **ith** integerof 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-negativeand no integer will be greater than **1000**.

# Output #

For each case, print the case number and the expected numberof gold you will collect. Errors less than **10\-6** will beignored.

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   <td> <h1>Sample Input</h1> </td> 
   <td> <h1>Output for Sample Input</h1> </td> 
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   <td> <p>3</p> <p>&nbsp;</p> <p>1</p> <p>101</p> <p>&nbsp;</p> <p>2</p> <p>10 3</p> <p>&nbsp;</p> <p>3</p> <p>3 6 9</p> </td> 
   <td> <p>Case 1: 101.0000000000</p> <p>Case 2: 13.000</p> <p>Case 3: 15</p> </td> 
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题意: 有n个格子, 每个格子有对应的金子数目, 从第一个格子开始,每次扔一个骰子, 骰子朝上的点数决定你向前走的步数, 如果超过n,那么一直扔, 直到小于n, 到最后到达n点, 求所得的期望金子的总数目.

分析: 求期望的话可以倒着来, 从n开始到1, dp\[i+j\](j>=1 && j<=6)可以转移到dp\[i\],当i+j<=n时, 扔骰子是等概率的, 因此dp\[i\] = dp\[i+j\]/6, 但是当i+j>n的时候就不是1/6的概率了,要看j的最大值能取到多少, 然后就是1/j;

#include<bitset>
    #include<map>
    #include<vector>
    #include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<cmath>
    #include<stack>
    #include<queue>
    #include<set>
    #define inf 0x3f3f3f3f
    #define mem(a,x) memset(a,x,sizeof(a))
    #define F first
    #define S second
    using namespace std;
    
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<int,int> pii;
    
    inline int in()
    {
        int res=0;char c;int f=1;
        while((c=getchar())<'0' || c>'9')if(c=='-')f=-1;
        while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
        return res*f;
    }
    const int N=100010,MOD=1e9+7;
    
    int a[123];
    double dp[123];
    int main()
    {
        int T=in(),ca=1;
        while(T--)
        {
            int n=in();
            for(int i=0;i<n;i++)
            {
                a[i]=in();
            }
            dp[n-1]=a[n-1];
            for(int i=n-2;i>=0;i--)
            {
                dp[i]=a[i];
                int k=0;
                for(int j=i+1;j<=i+6;j++)
                {
                    if(j==n) break;
                    k++;
                }
                for(int j=i+1;j<=i+6;j++)
                {
                    if(j<n)
                    {
                        dp[i]+=dp[j]*1.0/k;
                    }
                }
            }
            printf("Case %d: %.9lf\n",ca++,dp[0]);
        }
        return 0;
    }