tsinsen-a1489 lucky Draw "mathematical expectation"

Jominda too God, in fact, has been very clear, I will push it again.

The title means that there are N boxes, no difference cast m ball, each box score is the number of balls in each box.

First question:

Suppose the ball is in box I, then? ans = (mi + 1) ^ 2-mi ^ 2--ans = 2mi + 1

Also take expectations-E (? i) = 2E (? mi) + 1

I this box has an MI ball (assuming that a T-ball has been cast) for the T * pi to substitute E (? i) = 2 * t * pi + 1

Special to general, for any i,i = the probability of the current selected i is pi, then E (?) = Sigma (pi * E (? i)), substituting, E (?) = Sigma (2 * t * pi ^ 2 + pi) = 1 + 2 * t * Sigma (pi ^ 2).

So for t = [0, m-1], sum, ans = sigma (1 + 2 * t * Sigma (pi ^ 2)), ans = m + M * (m-1) * Sigma (pi ^ 2).

　　

Second question:

For a person, we consider whether his winning situation, define it as Ci,ci if 1 is winning, otherwise. The answer is E (Sigma (CI)) = Sigma (E (CI)).

E (CI) probability for a person to win, E (CI) = 1-(1-pi) ^ M.

1#include <bits/stdc++.h>2 #defineRep (I, A, b) for (int i = A; I <= b; i++)3 #defineDrep (I, A, b) for (int i = A; I >= b; i--)4 #defineREP (I, A, b) for (int i = A; I < b; i++)5 #defineMP Make_pair6 #definePB Push_back7 #defineCLR (x) memset (x, 0, sizeof (x))8 #defineXX First9 #defineyy secondTen using namespacestd; OnetypedefLong Longi64; Atypedef pair<int,int>PII; - Const intINF = ~0U>>1; - Consti64 INF = ~0ull >>1; the //******************************* - Const intMAXN =100005; - intC[MAXN]; - DoublePow (Double Base,intnum) { +     DoubleRET =1; -      while(num) { +         if(Num &1) RET *=Base; A         Base*=Base; atNum >>=1; -     } -     returnret; - } - intMain () { -     intN, M; inscanf"%d%d", &n, &m); -I64 sum (0); toRep (I,1, N) scanf ("%d", &c[i]), sum + =C[i]; +     DoubleANS1 (0); -     Static DoubleP[MAXN]; theRep (I,1, n) p[i] =1.0* C[i]/sum; *Rep (I,1, n) ans1 + = p[i] *P[i]; $Ans1 *=m;Panax NotoginsengAns1 *= M-1; -Ans1 + =m; theprintf"%.2lf\n", ans1); +     DoubleAns2 =N; ARep (I,1, n) ans2-= Pow (1.0-1.0*P[i], m); theprintf"%.2lf\n", ans2); +     return 0; -}

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tsinsen-a1489 lucky Draw "mathematical expectation"