lightoj 1030 Discovering Gold 期望

dp[i]表示，从i点出发到结束位置，期望获得的黄金。

dp[i] = val[i] + ∑dp[j] / cnt(j是i能一步到达的位置，cnt是一步能到达的位置的数量)

#include <cstdio>
#include <algorithm>
using namespace std;
int T,cas,n;
int val[105];
double dp[105];
int main()
{
    for (scanf("%d",&T);T != 0;T--)
    {
        cas++;
        scanf("%d",&n);
        for (int i = 1;i <= n;i++)
            scanf("%d",&val[i]);
        dp[n] = val[n];
        for (int i = n - 1;i >= 1;i--)
        {
            double sum = 0;
            int cnt = 0;
            for (int j = i + 1;j <= min(n,i + 6);j++)
            {
                sum += dp[j];
                cnt++;
            }
            dp[i] = val[i] + sum / cnt;
        }
        printf("Case %d: %.6lf
",cas,dp[1]);
    }
}