N1(8.3) solution

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The main idea: nn in the box at the beginning$n$ black balls$m$ white balls,ppeach timeThe probability of$p$ is put in a black ball, there is$1-$The probability of$p$ is to put a white ball in, and then randomly take a ball after the ball is placed, and ask forkkExpected number of black balls in the box after$k$ times.

answer

It can be noticed that the number of balls after each operation is maintained at $n+m$ will not change.

Ordinance $a_i$Means to proceed iiThe expected number of black balls after$i$ operations, obviously$a_0=n$。

There are four cases to consider, two of which are 放一个黑球拿一个黑球and 放一个白球拿一个白球, two balls will not cause changes in the number of black so no need to consider.

The probability of putting a black ball and picking a white ball is: $A=p\times \frac{n+m+1-(a_i+1)}{n+m+1}$, Contribution is $1$ , because there is an extra black ball.

The probability of putting a white ball and picking a black ball is: $B=(1-p)\times \frac{a_i}{n+m+1}$, The contribution is $-1$ because one black ball is missing.

整理得到：$a_{i+1}=a_i+A\times 1+B\times (-1)=(p+a_i)\frac{n+m}{n+m+1}$。

令$T=\frac{n+m}{n+m+1}$， Naka Yu $a_{i+1}=T{a}_i+pT$ , expand to get$a_i=T^i{a}_0+pT(1+T+T^2+...+T^{i-1})$, then$O(\log k)$ Calculate the answer.

code show as below:

#include <cstdio>
#define mod 1000000007

int n,m,k,a,b;
int ksm(int x,int y){
    
    int re=1;for(;(y&1?re=1ll*re*x%mod:0),y;x=1ll*x*x%mod,y>>=1);return re;}
#define inv(x) ksm(x,mod-2)

int main()
{
    
    
	scanf("%d %d %d %d %d",&n,&m,&k,&a,&b);
	int p=1ll*a*inv(b)%mod,T=1ll*(n+m)*inv(n+m+1)%mod;
	printf("%d",(1ll*ksm(T,k)*n%mod+1ll*p*T%mod*(ksm(T,k)-1+mod)%mod*inv(T-1)%mod)%mod);
}