LightOJ 1027 A Dangerous Maze 概率期望学习小记

我会带着你远行 2021-11-05 02:38 260阅读 0赞

偶尔做过一些概率期望的题，现在系统的学习下。大学《概率论与统计》的知识基本够用……

下面的链接是别人总结的题目：

[概率DP\_滴答\_百度空间][DP]

[【DP\_概率DP专辑】【10、4最新更新】 - ZeroClock - 博客频道 - CSDN.NET][DP_DP_10_4_ - ZeroClock - _ - CSDN.NET]

[概率dp 与 dp求期望 - 标签 - Because Of You - 博客园][dp _ dp_ - _ - Because Of You -]

[概率dp与dp求期望 « DarkSword][dp_dp_ _ DarkSword]

[概率dp - Qiuqiqiu - 博客园][dp - Qiuqiqiu -]

[概率题总汇 - A Crazy Man - C++博客][- A Crazy Man - C]

LightOJ 1027 A Dangerous Maze

题目链接：http://lightoj.com/volume\_showproblem.php?problem=1027

思路：

走出去时间的期望E = (T1 + T2 + T3) / 3

3E=a\[1\]+(-a\[2\]+E)+(-a\[3\]+E)

E=a\[1\]-a\[2\]-a\[3\]

#include <cstdio>
    
    int Gcd(int x,int y)  
    {  
        return !y?x:Gcd(y,x%y);  
    }  
    
    int main ()
    {
    	int T;
    	scanf("%d",&T);
    	for (int Cas=1;Cas<=T;Cas++)
    	{
    		int n,m=0,s=0,tmp;
    		scanf("%d",&n);
    		for (int i=1;i<=n;i++)
    		{
    			scanf("%d",&tmp);
    			if (tmp<0)
    				s-=tmp;
    			else
    				m++,s+=tmp;
    		}
    		if (m==0)  //全为负数
    			printf("Case %d: inf\n",Cas);
    		else
    			printf("Case %d: %d/%d\n",Cas,s/Gcd(m,s),m/Gcd(m,s));
    	}
    	return 0;
    }

[DP]: http://hi.baidu.com/sunnydaughter/item/e2d5b2350a61e295b80c03e7
[DP_DP_10_4_ - ZeroClock - _ - CSDN.NET]: http://blog.csdn.net/woshi250hua/article/details/7912049
[dp _ dp_ - _ - Because Of You -]: http://www.cnblogs.com/wuyiqi/tag/%E6%A6%82%E7%8E%87dp%20%E4%B8%8E%20dp%E6%B1%82%E6%9C%9F%E6%9C%9B/
[dp_dp_ _ DarkSword]: http://www.ilovedou.com/?cat=39
[dp - Qiuqiqiu -]: http://www.cnblogs.com/yangcl/archive/2013/01/17/2865573.html
[- A Crazy Man - C]: http://www.cppblog.com/notonlysuccess/archive/2009/05/19/83367.html