传送门
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
分析
我们令f[i]为结束概率
[f[i]=C_{i-1}^{k-1}*p^k*(1-p)^{i-k}
]
[sum f[i]=1(关键）
]
(ans1=sum f[i])
(=sum i*C_{i-1}^{k-1}*p^k*(1-p)^{i-k})
(=p^k*k*sum C_i^k*(1-p)^{i-k})
(=k/p)
(ans2=sum i*i*f[i])
(=k*p^k*sum i*C_i^k*(1-p)^{i-k})
(=k*p^k*sum C_{i+1}^{k+1}*(1-p)^{i-k}-ans1)
(=frac {k*(k+1)}{p^2}-ans1)
trick
代码
#include<cstdio>
#include<cstring>
double k,p;
const double eps=1e-10;
int main()
{
while(scanf("%lf",&k)!=EOF)
{
if(k<eps) break;
scanf("%lf",&p);
printf("%.3f %.3f
",k/p,(k+1)*k/(p*p)-k/p);
}
}