zoj3329 One Person Game

最新推荐文章于 2021-04-10 21:56:02 发布

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最新推荐文章于 2021-04-10 21:56:02 发布

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分类专栏： zoj 概率论数学

本文链接：https://blog.youkuaiyun.com/qq_41510496/article/details/79394032

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本文介绍了一种计算一人数独游戏中投掷骰子次数期望值的方法。玩家使用三个不同面数的骰子进行游戏，根据特定规则更新计数器数值，直至达到预设值n结束游戏。通过数学和编程手段，求解在此过程中玩家需要投掷骰子的平均次数。

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One Person Game

Time Limit: 1 Second Memory Limit: 32768 KB Special Judge

There is a very simple and interesting one-person game. You have 3 dice, namely Die1, Die2 and Die3. Die1 has K₁ faces. Die2 has K₂ faces. Die3 has K₃ faces. All the dice are fair dice, so the probability of rolling each value, 1 to K₁, K₂, K₃ is exactly 1 / K₁, 1 / K₂ and 1 / K₃. You have a counter, and the game is played as follow:

Set the counter to 0 at first.
Roll the 3 dice simultaneously. If the up-facing number of Die1 is a, the up-facing number of Die2 is b and the up-facing number of Die3 is c, set the counter to 0. Otherwise, add the counter by the total value of the 3 up-facing numbers.
If the counter's number is still not greater than n, go to step 2. Otherwise the game is ended.

Calculate the expectation of the number of times that you cast dice before the end of the game.

Input

There are multiple test cases. The first line of input is an integer T (0 < T <= 300) indicating the number of test cases. Then T test cases follow. Each test case is a line contains 7 non-negative integers n, K₁, K₂, K₃, a, b, c (0 <= n <= 500, 1 < K₁, K₂, K₃ <= 6, 1 <= a <= K₁, 1 <= b <= K₂, 1 <= c <= K₃).

Output

For each test case, output the answer in a single line. A relative error of 1e-8 will be accepted.

Sample Input

2
0 2 2 2 1 1 1
0 6 6 6 1 1 1

Sample Output

1.142857142857143
1.004651162790698

Author: CAO, Peng

Source: The 7th Zhejiang Provincial Collegiate Programming Contest

题解：这种题目我自己做出来还是想都别想了orz

吐个槽，怎么学个oi还扯那么多数学

大牛：https://www.cnblogs.com/kuangbin/archive/2012/10/03/2710648.html

代码：

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int main(){

    //freopen("input.txt","r",stdin);

    int t,n,k1,k2,k3,a,b,c;
    scanf("%d",&t);
    double p0,p[20];
    while(t--){
        cin>>n>>k1>>k2>>k3>>a>>b>>c;
        int tot=k1+k2+k3;
        p0=1.0/(k1*k2*k3);
        memset(p,0,sizeof(p));
        for(int i=1;i<=k1;i++)
            for(int j=1;j<=k2;j++)
                for(int k=1;k<=k3;k++)
                    if(i!=a || j!=b || k!=c)
                        p[i+j+k]+=p0;
        double a[520]={0},b[520]={0};
        for(int i=n;i>=0;i--){
            for(int k=3;k<=tot;k++){
                a[i]+=a[i+k]*p[k];
                b[i]+=b[i+k]*p[k];
            }
            a[i]+=p0;
            b[i]+=1;
        }
        printf("%.15lf\n",b[0]/(1-a[0]));
    }
    return 0;
}