Card（bestcoder #26 B)

 

 

Card

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 218    Accepted Submission(s): 86
Special Judge

Problem Description

There are x cards on the desk, they are numbered from 1 to x. The score of the card which is numbered i(1<=i<=x) is i. Every round BieBie picks one card out of the x cards，then puts it back. He does the same operation for b rounds. Assume that the score of the j-th card he picks is S j   . You are expected to calculate the expectation of the sum of the different score he picks.

 

Input

Multi test cases，the first line of the input is a number T which indicates the number of test cases.
In the next T lines, every line contain x,b separated by exactly one space.

[Technique specification]
All numbers are integers.
1<=T<=500000
1<=x<=100000
1<=b<=5

 

Output

Each case occupies one line. The output format is Case #id: ans, here id is the data number which starts from 1，ans is the expectation, accurate to 3 decimal places.
See the sample for more details.

 

Sample Input

2 2 3 3 3

 

Sample Output

Case #1: 2.625 Case #2: 4.222

Hint

For the first case, all possible combinations BieBie can pick are (1, 1, 1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2) For (1,1,1),there is only one kind number i.e. 1, so the sum of different score is 1. However, for (1,2,1), there are two kind numbers i.e. 1 and 2, so the sum of different score is 1+2=3. So the sums of different score to corresponding combination are 1，3，3，3，3，3，3，2 So the expectation is (1+3+3+3+3+3+3+2)/8=2.625

 

 

 

1002 Card
设Xi代表分数为i的牌在b次操作中是否被选到,Xi=1为选到，Xi=0为未选到
那么期望EX=1*X1+2*X2+3*X3+…+x*Xx
Xi在b次中被选到的概率是1-(1-1/x)^b
那么E(Xi)= 1-(1-1/x)^b
那么EX=1*E(X1)+2*E(X2)+3*E(X3)+…+x*E(Xx)=(1+x)*x/2*(1-(1-1/x)^b)

转载于:https://www.cnblogs.com/get-an-AC-everyday/p/4219055.html