lightoj 1030 水概率

最新推荐文章于 2021-06-22 21:54:43 发布

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你身处一个1xN的洞穴，每个位置可能含有不同数量的黄金。从位置1开始，每次投掷六面骰子，前进相应步数并收集黄金。若超出洞穴范围，则继续投掷直至进入有效位置。当到达N号位置时结束。给定洞穴信息，计算按此规则能收集到的预期黄金数。输入包含测试用例数T及每个洞穴的长度N和黄金分布，输出每个案例的预期黄金数。

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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 Nth 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
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 ith integer of this line denotes the amount of gold you will get if you come to the ith 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
3

1
101

2
10 3

3
3 6 9
Case 1: 101.0000000000
Case 2: 13.000
Case 3: 15

就是个概率….

#include<iostream>
#include<cmath>
#include<string>
#include<algorithm>
#include<memory.h>
#include<cstdio>
using namespace std;
double dp[101];
double tu[101];
int main()
{
    int T;
    cin>>T;
    int u=0;
    while(T--)
    {
        int n;
        cin>>n;
        memset(dp,0,sizeof(dp));
        for(int a=1;a<=n;a++)cin>>tu[a];
        dp[n]=tu[n];
        for(int a=n-1;a>=1;a--)
        {
            int shengyu;
            if(n-a>6)shengyu=6;
            else shengyu=n-a;
            dp[a]+=tu[a];
            for(int b=a+1;b<=a+shengyu;b++)dp[a]+=dp[b]/shengyu;
        }
        printf("Case %d: %.9f\n",++u,dp[1]);
    }
    return 0;
}