题干：
Ilya got tired of sports programming, left university and got a job in the subway. He was given the task to determine the escalator load factor.
Let's assume that n people stand in the queue for the escalator. At each second one of the two following possibilities takes place: either the first person in the queue enters the escalator with probability p, or the first person in the queue doesn't move with probability (1 - p), paralyzed by his fear of escalators and making the whole queue wait behind him.
Formally speaking, the i-th person in the queue cannot enter the escalator until people with indices from 1 to i - 1 inclusive enter it. In one second only one person can enter the escalator. The escalator is infinite, so if a person enters it, he never leaves it, that is he will be standing on the escalator at any following second. Ilya needs to count the expected value of the number of people standing on the escalator after t seconds.
Your task is to help him solve this complicated task.
Input
The first line of the input contains three numbers n, p, t (1 ≤ n, t ≤ 2000, 0 ≤ p ≤ 1). Numbers n and t are integers, number p is real, given with exactly two digits after the decimal point.
Output
Print a single real number — the expected number of people who will be standing on the escalator after t seconds. The absolute or relative error mustn't exceed 10 - 6.
Examples
Input
1 0.50 1
Output
0.5
Input
1 0.50 4
Output
0.9375
Input
4 0.20 2
Output
0.4题目大意：
n个人在排成一队在电梯面前，最前面的人每一秒钟会进行一次选择，有p的概率进电梯，有1-p的概率停在原地。每个人只有他前面的人都进电梯了，他才有可能进电梯。求t秒之后，进电梯人数的期望值。
解题报告：
看得出来，还是对概率dp的理解不是很透彻，换句话说是对dp的理解不是很透彻。转移的时候别忘记乘以对应的概率。
dp[i][j]代表i秒过后，电梯上有j个人的概率 。这样就可以由两个状态转移而来，但是注意这时候要注意这只是定义了某个状态发生的概率，转移的同时还要乘以转移的概率，这两个概率不是一个东西，所以不能直接叠加上来。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e3 + 5;
int n,t;
double p,dp[MAX][MAX];//i秒过后，电梯上有j个人的概率
int main()
{cin>>n>>p>>t;dp[0][0]=1;for(int i = 1; i<=t; i++) {dp[i][0] = dp[i-1][0]*(1-p);for(int j = 1; j<=n; j++) {dp[i][j] = dp[i-1][j] * (j == n ? 1 : (1-p));dp[i][j] += dp[i-1][j-1] * p;}}double ans = 0;for(int i = 1; i<=n; i++) ans += dp[t][i]*i;printf("%.8f\n",ans);return 0 ;
}注意状态的转移不能直接叠加上来，比如说这样：要么第i秒没有人上来，那么就是i-1秒上来了j个人的概率；要么就是上来了一个人，就要乘以对应的概率。
for(int i = 1; i<=t; i++) {dp[i][0] = dp[i-1][0]*(1-p);for(int j = 1; j<=n; j++) {dp[i][j] = dp[i-1][j];dp[i][j] += dp[i-1][j-1] * p;}}甚至还写出了这种四不像：
// for(int i = 1; i<=n; i++) {
// for(int j = 1; j<=t; j++) {
// dp[i][j] += dp[i][j-1];
// dp[i][j] += (1-dp[i][j-1])*p;
// }
// }
总结：
其实就BZOJ - 3036这个算 经过路径长度的期望 一样：
但是那代码就要这么写了：
using namespace std;
int n, t;
double p, dp[2005][2005];int main() {scanf("%d%lf%d", &n, &p, &t);dp[0][0] = 1;double ans = 0;for (int i = 0; i < t; ++i) {dp[i + 1][n] += dp[i][n];for (int j = 0; j < n; ++j) if (dp[i][j] > 1e-10) {dp[i + 1][j + 1] += dp[i][j] * p;dp[i + 1][j] += dp[i][j] * (1 - p);ans += dp[i][j] * p;}}printf("%.8f\n", ans);return 0;
}
