lightoj 1030 - Discovering Gold 【期望】

最新推荐文章于 2019-04-02 18:50:51 发布

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你身处一个可以用1xN网格表示的洞穴，每个格子可能含有不同数量的黄金。从位置1开始，每次扔一枚六面骰子，根据掷出的点数前进并收集相应位置的黄金。如果到达洞穴末端N，则停止探索。给定洞穴的黄金分布，你需要计算按指定规则探索的预期黄金总和。输入包含测试用例数T和每个洞穴的尺寸N，接下来是N个表示各格黄金数量的整数。输出每个案例的编号和预期收集的黄金数量，误差小于10^-6会被忽略。

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1030 - Discovering Gold

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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^thposition 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^thinteger 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

Output for Sample Input

101

10 3

3 6 9

Case 1: 101.0000000000

Case 2: 13.000

Case 3: 15

PROBLEM SETTER: JANE ALAM JAN

题意：给定n个格子以及每个格子上的gold值，你从第一个格子出发，每次掷1-6的骰子，根据掷出的值前进。若当前位置 + 掷出的值 > n，则重新掷骰子，到达第n个格子游戏结束。问你获得gold值的期望。

思路：设dp[i]为到达i个格子获得gold值的期望，则有dp[i] = (dp[i+1]/6 + ... + dp[i+6]/6) + dp[i]。

初始化dp[i] = 第i个格子的gold值。

AC代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <vector>
#define INF 0x3f3f3f3f
#define eps 1e-8
#define MAXN (3000+10)
#define MAXM (50000000)
#define Ri(a) scanf("%d", &a)
#define Rl(a) scanf("%lld", &a)
#define Rf(a) scanf("%lf", &a)
#define Rs(a) scanf("%s", a)
#define Pi(a) printf("%d\n", (a))
#define Pf(a) printf("%lf\n", (a))
#define Pl(a) printf("%lld\n", (a))
#define Ps(a) printf("%s\n", (a))
#define W(a) while(a--)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define MOD 1000000007
#define LL long long
#define lson o<<1, l, mid
#define rson o<<1|1, mid+1, r
#define ll o<<1
#define rr o<<1|1
using namespace std;
int gcd(int a, int b){
    return b == 0 ? a : gcd(b, a%b);
}
double dp[110], num[110];
int main()
{
    int t, kcase = 1; Ri(t);
    W(t)
    {
        int n; Ri(n);
        for(int i = 0; i < n; i++)
            Rf(num[i]), dp[i] = num[i];
        for(int i = n-2; i >= 0; i--)
        {
            int choose = min(6, n - i - 1);
            for(int j = 1; j <= choose; j++)
                dp[i] += dp[i+j] / (double)choose;
        }
        printf("Case %d: %.7lf\n", kcase++, dp[0]);
    }
    return 0;
}