https://codeforces.com/problemset/problem/518/D
Ideas:
Define dp[t][j] as the probability that the person j gets on the elevator by t seconds
Transfer:
dp[t][j]=dp[t-1][j-1]*p+dp[t-1][j]*(1-p);
Boundary: when j==0, the transfer still needs *(1-p)
When j==n, the transfer of the former is 100%
The answer is ans+=dp[t][j:0~n]*j
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=2e3+100;
typedef long long LL;
inline LL read(){LL x=0,f=1;char ch=getchar(); while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
double dp[maxn][maxn];///到t秒 j个人上了电梯的概率
int main(void)
{
///cin.tie(0);std::ios::sync_with_stdio(false);
LL n,t;double p;
n=read();
scanf("%lf",&p);
t=read();
dp[1][0]=1.0-p;
dp[1][1]=1.0*p;
for(LL i=2;i<=t;i++){///枚举时间
for(LL j=0;j<=n;j++){///枚举人
if(j==0) dp[i][j]=dp[i-1][j]*(1-p);
else if(j==n) dp[i][j]=dp[i-1][j]+dp[i-1][j-1]*p;
else dp[i][j]=dp[i-1][j-1]*p+dp[i-1][j]*(1.0-p);
}
}
double ans=0;
for(LL j=0;j<=n;j++){
ans+=1.0*j*dp[t][j];
}
printf("%.8f\n",ans);
return 0;
}