lightoj 1030 Discovering Gold[ 期望 ]_you are in a cave, a long cave! the cave can be re-优快云博客

本文链接：https://blog.youkuaiyun.com/code_or_code/article/details/45489143

本文深入探讨了在一个由洞穴构成的网格中寻找黄金的数学问题，通过模拟掷骰子的过程，计算到达特定位置时收集到的黄金数量的期望值。通过多种解题姿势，包括递归、动态规划和概率计算，读者可以理解如何运用数学原理解决实际问题。

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B - Discovering Gold

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 N^th 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 i^th integer of this line denotes the amount of gold you will get if you come to the i^th 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

101

10 3

3 6 9

Sample Output

Case 1: 101.0000000000

Case 2: 13.000

Case 3: 15

题目大意：从1位置摇骰子，到n位置，每个位置有权值，求最后得到的权值期望；

代码：

姿势一：YY乱搞；

代码：

int v[110];
double dp[110][110];
double f[110];

int main()
{
    int T,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",v+i);
        for(int i=0;i<=n;i++){
            f[i]=0.0;
            for(int j=0;j<=n;j++)
                dp[i][j]=0.0;
        }

        dp[1][1]=1.0;
        f[1]=1.0;
        int flag;

        for(int i=2;i<=n;i++){
            for(int j=1;j<=n-1;j++){
                flag=0;
                for(int k=6;k>=1;k--){
                    if(!flag&&j+k<=n){
                        flag=k;
                    }
                    if(flag){
                        dp[i][j+k]+=dp[i-1][j]/flag;

                    }
                }
            }
            for(int j=1;j<=n;j++)
                f[j]+=dp[i][j];
        }
        double ans=0.0;
        for(int i=1;i<=n;i++){
           // cout<<"-"<<f[i]<<endl;
            ans+=f[i]*v[i];
        }
        printf("Case %d: %.10lf\n",++cas,ans);
    }

    return 0;
}

姿势二：dp[i]表示重i到n得到权值的期望 dp[i]=v[i]+dp[i+k]/j;

代码：

int v[110];
double dp[110];

int main()
{
    int T,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",v+i);

        dp[n]=v[n];
        for(int i=n-1;i>=1;i--){
            int k=min(6,n-i);
            dp[i]=v[i];
            for(int j=1;j<=k;j++){
                dp[i]+=dp[i+j]/k;
            }
        }

        printf("Case %d: %.10f\n",++cas,dp[1]);
    }

    return 0;
}

姿势三：dp[i] 经过i位置的概率，最后求ans;

代码：

int v[110];
double dp[110];

int main()
{
    int T,cas=0;
    scanf("%d",&T);
    while(T--)
    {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            dp[i]=0.0;
            scanf("%d",v+i);
        }

        dp[1]=1;
        for(int i=1;i<=n-1;i++){
            int k=min(6,n-i);
            for(int j=1;j<=k;j++)
                dp[i+j]+=dp[i]/k;
        }

        double ans=0.0;
        for(int i=1;i<=n;i++)
            ans+=dp[i]*v[i];
        printf("Case %d: %.10f\n",++cas,ans);
    }

    return 0;
}