一、原题
You are in a maze; seeing n doors in front of you in beginning. You can choose any door you like. The probability for choosing a door is equal for all doors.
If you choose the ith door, it can either take you back to the same position where you begun in xi minutes, or can take you out of the maze after ximinutes. If you come back to the same position, you can't remember anything. So, every time you come to the beginning position, you have no past experience.
Now you want to find the expected time to get out of the maze.
InputInput 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 number of doors. The next line contains n space separated integers. If the ithinteger (xi) is positive, you can assume that the ith door will take you out of maze after xi minutes. If it's negative, then the ith door will take you back to the beginning position after abs(xi) minutes. You can safely assume that 1 ≤ abs(xi) ≤ 10000.
OutputFor each case, print the case number and the expected time to get out of the maze. If it's impossible to get out of the maze, print 'inf'. Print the result in p/q format. Where p is the numerator of the result and q is the denominator of the result and they are relatively prime. See the samples for details.
Sample Input3
1
1
2
-10 -3
3
3 -6 -9
Sample Output
Case 1: 1/1
Case 2: inf
Case 3: 18/1
二、题目大意
(翻译来自互联网)
三、分析
设ai为可以出去的门的花费,bi为不可以出去的门的花费,sum()为求和函数,e为从迷宫出去的期望(花费),numb为b类门的个数。因为有n个门,又是等概率选的且每次选择相互独立,所以每个门被选择的概率:p=1/n
则
- 一次可以出去的花费为:sum(ai*p)=p*sum(ai):直接套用概率论的公式:一个事件的期望=每种情况的期望的期望 * 对应的概率
- 一次不可以出去的花费为:p*sum(bi+e)=p*sum(bi)+numb*e意思是 花费bi+从相同的迷宫出去的花费,进一步解释:即用了一个门之后,又回到了原点,而从原点出去的期望为e,所以总花费是bi+e,而对于每个不能出去的门都有相同的结论,所以直接求和sum(bi+e)。ps:此处e也是所求的期望,此处假设e是已知的。
综上所述,所求的e为:
e=以上两种情况之和=p*sum(ai)+p*sum(bi)+p*numb*e;
移项,将e移到左边。
(1-p*numb)*e=p*[sum(ai)+sum(bi)]
化简得
e=p/(1-p*numb)*[sum(ai)+sum(bi)]
然后p=1/n,代入化简得
e=[sum(ai)+sum(bi)] / (n-numb)
推导完毕。
然后再特判一下inf就好了。
四、AC代码
仅供参考(第一次写的时候思路不太清楚,变量名写的比较乱,凑合着看吧)
#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
typedef long long ll;
int main()
{
int n,T;
cin>>T;
int i,t;
int sumtc,sumts;//c为可以出去的,s为不可以
int c,s;
int cs=1;
while(T--)
{
cin>>n;
sumtc=sumts=s=c=0;
for(i=0;i<n;i++)
{
scanf("%d",&t);
if(t>0)
{
sumtc+=t;
c++;
}
else if(t<0)
{
sumts+=-t;
s++;
}
}
if(!c)
{
printf("Case %d: inf\n",cs++);
continue;
}
int fz=sumtc+sumts;
int fm=n-s;
int b=__gcd(fz,fm);
fz/=b;
fm/=b;
printf("Case %d: %d/%d\n",cs++,fz,fm);
}
}