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最新推荐文章于 2019-08-16 10:46:51 发布

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最新推荐文章于 2019-08-16 10:46:51 发布

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分类专栏： ACM_C语言

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本文介绍了一个有趣的算法问题——如何计算在一个带有黄金的线性路径中，通过掷骰子前进并收集黄金的期望收益。文章提供了一段C++代码实现，详细解释了动态规划方法的应用。

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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 expectednumber 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

题意：

有一个直线的金矿，每个点有一定数量的金子；

你从０开始，每次扔个骰子，扔出几点就走几步，然后把那个点的金子拿走；

如果扔出的骰子超出了金矿，就重新扔，知道你站在最后一个点；

问拿走金子的期望值是多少；

首先我们假设你现在站在第ｉ个点，且从这个点开始走；

那么这个点的期望p[i] = p[i +1] /6 + p[i + 2] / 6 + p[i + 3] /6 + p[i + 4] / 6 + p[i + 5] / 6 + p[i + 6] / 6 + p[i]；

p[i]　初值就是这个点的金子数量，意思就是这个点的期望，是往后有６种情况，每种的六分之一；

当然情况数少于６的时候要处理一下；

如果小于6，那么每次概率都会随着你每一次用掉之后而增加，（倒着来算）加入是4，开始就是除1， 2， 3，

大于6 那就是6了，实际还有点不懂

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<algorithm>
#include<iostream>
using namespace std;
#define N 1000
double a[110], f[110];
int main()
{
int i, j, n, T, t=1;
scanf("%d", &T);
while(T--)
{
scanf("%d", &n);
for(i=0;i<n;i++)
{
scanf("%lf", &a[i]);
}
f[n-1]=a[n-1];
for(i=n-2;i>=0;i--)
{
f[i]=a[i];
int dis=6;
if(n-1-i<6)
dis=n-1-i;

for(j=1;j<=dis;j++)
{
f[i]+=(f[i+j]/dis);
}
}
printf("Case %d: %.10f\n", t++, f[0]);
}
return 0;
}