HDU 5159 Card（数学期望）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5159

Card

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 Sj .
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
HintFor 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

题意：

X张卡牌，标号分别为1，2，…,X,,每张牌的分数为对应的序号，随机抽取卡牌b次，每次都放回，记录下卡牌的分数，求抽取到不同卡牌的期望

思路：

和的期望
= 各部分期望的和。

E = E(1) + E(2) + ... + E(x) ;

每张牌每一轮抽到的概率为 1/x ;

每张牌每一轮不抽到的概率为 1-1/x ；

每张牌b次操作后都没抽到的概率为 （1-1/x)^b ;

每张牌b次操作后有被抽到的概率为 p = 1-（1-1/x)^b ;

所以每个数的期望也是 i*( 1-(1-1/x)^b )

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int main()
{
    int t,k=1;
    scanf("%d",&t);
    double x,b;
    while(t--)
    {
        scanf("%lf%lf",&x,&b);
        double ans=0;
        double p=1-pow((1-1.0/x),b);
        double n=(1+x)*x*1.0/2;
        ans=n*p;
        printf("Case #%d: %.3lf\n",k++,ans);
    }
    return 0;
}