学步园

题意：进行翻硬币实验，若k次向上则结束，进行第n次实验需花费2*n-1的费用，询问期望结束次数及期望结束费用

设f[i]为结束概率

f[i]=c(i-1,k-1)*p^k*(1-p)^(i-k)

sigma(f[i])==1

p^k*sigma(c(i-1,k-1)*(1-p)^(i-k))=1

sigma(c(i-1,k-1)*(1-p)^(i-k))=1/(p^k)

ans1=sigma(i*f[i])

=p^k*sigma(i*c(i-1,k-1)*(1-p)^(i-k)) //将i放入组合数

=k*p^k*sigma(c(i,k)*(1-p)^(i-k))

=k*p^k*p^(k+1)

=k/p

ans2=sigma(i*i*f[i])

=p^k*sigma(i*i*c(i-1,k-1)*(1-p)^(i-k))

=k*p^k*sigma(i*c(i,k)*(1-p)^(i-k))

=k*p^k*sigma((i+1)*c(i,k)*(1-p)^(i-k))-p^k*sigma(c(i,k)*(1-p)^(1-k))

=k*(k+1)*p^k*sigma(c(i+1,k+1)*(1-p)^(i-k))-ans1 //将i+1放进去

=k*(k+1)*p^k/(p^(k+2))-ans1

=k*(k+1)/p^2-ans1

=[(k+1)/p]*ans1-ans1

关键在于第一步利用概率和为1的性质，将复杂的带组合公式的无穷求和式化为较简单的有穷式

#include <cstdio>
#include <cstdlib>
#include <cstring>
int k;
double ans1,ans2,p;
void init()
{
    ans1=k/p;
    ans2=((double)k+1)/p*ans1-ans1;
    printf("%.3lf %.3lf\n",ans1,ans2);
}
int main()
{
    freopen("poj3682.in","r",stdin);
    freopen("poj3682.out","w",stdout);
    for (;;) {
	scanf("%d%lf\n",&k,&p);
	if (!k) break;
	init();
    }
    return 0;
}