The question ：

There are... On the ground n Planting sticks , There are two types , One type is recognizable , One type is unrecognizable , Each stick has a weight .

When you find something recognizable , Then you won't pick up this stick in the future , If it's not recognizable , Then you might pick up .

Ask the expectation that all the sticks will be collected .

Ideas ：

This topic draws lessons from this article ：Aladdin and the Magical Sticks

First , At first, it seems that , and LightOj 1027 A Dangerous Maze It's kind of like , It's just , This is to go through all the doors .

Let's start with a classic question ：

Stamp collection issues （Coupon Collector Problem）WiKi Information

The proof process of expectation of geometric distribution

When solving the stamp collection problem , We need to use geometric distribution expectation when we calculate expectation from probability , So the proof process of expectation of geometric distribution is given here . It's very concise , There are a lot of examples to understand .

Through the questions above , We can assume that , We are now facing a n Face dice , Every side of the dice is random （ It's like an unrecognizable stick ）, Ask for the expected number of times that all faces will be seen （ Let's just look at the top side ）

that , The solution to the problem is ：

H[n] = (1 + 1/2 + 1/3 + 1/4 + ... + 1/n), This is it. The first part of harmonic series n term .

This value is approximately equal to Euler constant about ：0.57721566490153286060651209.（ But it's a time when n An approximation to infinity , It's not a substitute for concrete H[n], For example, when n = 1 || 2 when ）

And what we're looking for is the expected weight , According to the linear property of expectation E(XY) = E(X)*E(Y)

therefore , The total weight expectation is equivalent to The weight expectation of each time * The number of expectations .

n Face to face , The expected number of times each face appears at least once is ：E(x) = n * H[n], that , The expected number of times a given face appears at least once is E(z) = E(x)/n = H[n].

therefore , Assume that this n When all sticks are unrecognizable, the expected weight is ：

Ea = E(w) * E(x), E(w) For the expectation of weight = The average of the weights .

however , this n Some of the sticks are recognizable , So subtract the extra expectations .

Let's first calculate the expected number of times a recognizable stick needs to be recognized , The answer is 1.

When there are six balls in the box , Using no return sampling , How many times do you expect to draw six balls ？ This is a fixed value , by 6.

therefore , The extra part of every stick is （H[n] - 1） * w[i].w[i] For the weight of some recognizable stick .

set up , The average weight of all sticks is Wn

Suppose there is k A recognizable stick , The average value of its weight is Wk

So , The answer for ： Ea - Eb = Wn * n * H[n] - k * Wk * (H[n] - 1)

Simplification ： E = (Wn * n - k * Wk) * H[n] + k * Wk.

Code ：

#include <cmath>

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <ctime>

#include <set>

#include <map>

#include <list>

#include <queue>

#include <string>

#include <vector>

#include <fstream>

#include <iterator>

#include <iostream>

#include <algorithm>

using namespace std;

#define LL long long

#define INF 0x3f3f3f3f

#define MOD 1000000007

#define eps 1e-6

#define MAXN 5050

#define MAXM 100

#define dd cout<<"debug"<<endl

#define pa {system("pause");}

#define p(x) printf("%d\n", x)

#define pd(x) printf("%.7lf\n", x)

#define k(x) printf("Case %d: ", ++x)

#define s(x) scanf("%d", &x)

#define sd(x) scanf("%lf", &x)

#define mes(x, d) memset(x, d, sizeof(x))

#define do(i, x) for(i = 0; i < x; i ++)

#define dod(i, x, l) for(i = x; i >= l; i --)

#define doe(i, x) for(i = 1; i <= x; i ++)

int n;

double h[MAXN];

void init()

{

h[] = ;

for(int i = ; i < MAXN; i ++)

h[i] = h[i - ] + 1.0 / i;

} int main()

{

int T;

int kcase = ;

init();

scanf("%d", &T);

while(T --)

{

scanf("%d", &n);

int a, b;

double ans = ;

for(int i = ; i < n; i ++)

{

scanf("%d %d", &a, &b);

ans += a * (b == ? : h[n]);

}

printf("Case %d: %.5lf\n", ++ kcase, ans);

}

return ;

}

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