LightOJ - 1027 A Dangerous Maze —— 期望

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发布时间：2019-06-14

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题目链接：

1027 - A Dangerous Maze

Time Limit: 2 second(s)

Memory Limit: 32 MB

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 i^th door, it can either take you back to the same position where you begun in x_i minutes, or can take you out of the maze after x_i minutes. 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.

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 number of doors. The next line contains n space separated integers. If the i^th integer (x_i) is positive, you can assume that the i^thdoor will take you out of maze after x_i minutes. If it's negative, then the i^th door will take you back to the beginning position after abs(x_i) minutes. You can safely assume that 1 ≤ abs(x_i) ≤ 10000.

Output

For 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 Input

Output for Sample Input

-10 -3

3 -6 -9

Case 1: 1/1

Case 2: inf

Case 3: 18/1

题意：

有n扇门，一扇门要么能在特定时间内把人带出迷宫，要么能在特定时间内把人带会原地，并使人失去记忆（即不知道这扇门是不能带出迷宫的），每扇门被选择的几率是相等的。问走出迷宫的平均时间。

题解：

设期望为EX，有t1扇“正门”，t2扇“负门”，其中t1+t2 = n 。第i扇门所花费的时间为ai

1.当选择的门为正门时，它可以直接走出迷宫，因此：1/n*ai，1/n为选择这扇门的几率，且所有的门被选择的几率也为1/n。

2.当选择的门为负门时，它先花费了ai的时间，然后又重新选择，重新选择然能走出迷宫的时间即为平均时间，因此：1/n*(ai+EX)。

3.综上，EX = ∑1/n*ai + ∑1/n*(ai+EX)，其中第一个i的范围为：1<=i<=t1，第二个i：1<=i<=t2。

代码如下：

1 #include 
      
        2 #include 
       
         3 #include 
        
          4 #include 
         
           5 #include 
          
            6 #include 
           
             7 #include 
            
              8 #include 
             
               9 #include 
              10 #include 
               
                11 #include 
                
                 12 using namespace std;13 typedef long long LL;14 const int INF = 2e9;15 const LL LNF = 9e18;16 const int MOD = 1e9+7;17 const int MAXN = 1e6+100;18 19 int gcd(int a, int b)20 {21 return b==0?a:gcd(b,a%b);22 }23 24 int main()25 {26 int T, n, kase = 0;27 scanf("%d", &T);28 while(T--)29 {30 scanf("%d", &n);31 int numP = 0, sum = 0;32 for(int i = 1; i<=n; i++)33 {34 int val;35 scanf("%d", &val);36 sum += abs(val);37 if(val>0) numP++;38 }39 if(numP==0)40 printf("Case %d: inf\n", ++kase);41 else42 printf("Case %d: %d/%d\n", ++kase, sum/gcd(sum, numP), numP/gcd(sum, numP));43 }44 }

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转载于:https://www.cnblogs.com/DOLFAMINGO/p/8442427.html

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