lightoj 1030 Discovering Gold[ 期望 ]

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 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.

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

3

1

101

2

10 3

3

3 6 9

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;
}

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;
}