LightOJ 1030 Discovering Gold(期望 概率)

正推，到达i的概率为p[i]，要注意除了1和n外，到达i的概率并不一定为1

概率表达式为p[i] += p[j] / min(n - j, 6)

从j带过来的期望为exp[i] += exp[j] / min(n - j, 6)

又到达i时有价值val[i]，到达i的概率为p[i],故exp[i] += val[i] * p[i]

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 1008, INF = 0x3F3F3F3F;
#define MS(a, num) memset(a, num, sizeof(a))
#define PB(A) push_back(A)
#define FOR(i, n) for(int i = 0; i < n; i++)
double exp[N], p[N];
int val[N];
int main(){
    int t;
    cin>>t;
    for(int cas= 1; cas <= t;cas++){
        int n;
        cin>>n;
        for(int i =1; i <= n; i++){
            scanf("%d", &val[i]);
        }
        p[1] = 1;
        exp[1] = val[1];
        for(int i = 2; i <= n; i++){
            exp[i] = p[i] = 0;
            for(int j = max(1, i - 6); j < i; j++){
                exp[i] += exp[j] / min(n - j, 6);
                p[i] += p[j] / min(n - j, 6);
            }
            exp[i] += val[i] * p[i];
        }
        printf("Case %d: %.10f\n", cas, exp[n]);
    }
    return 0;
}