Ilya and Escalator time limit per test: 2 seconds memory limit per test: 256 megabytes
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 from1 toi?-?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 aftert seconds.
Your task is to help him solve this complicated task.
Input
The first line of the input contains three numbersn,?p,?t (1?≤?n,?t?≤?2000,0?≤?p?≤?1). Numbers n and t are integers, numberp 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 aftert seconds. The absolute or relative error mustn't exceed10?-?6.
Sample test(s) Input
1 0.50 1
Output
0.5
Input
1 0.50 4
Output
0.9375
Input
4 0.20 2
Output
0.4
题目链接：http://codeforces.com/contest/518/problem/d
题目大意：n个人排队上电梯，排头每秒上去的概率为p，一共t秒，求t秒都电梯内人数的期望
题目分析：简单的概率dp，dp[i][j]表示第i秒电梯上有j个人的概率，最后累计一下期望
#include
#include
double dp[2005][2005]; int main() { int n, t; double p, ans = 0; memset(dp, 0, sizeof(dp)); dp[0][0] = 1; scanf("%d %lf %d", &n, &p, &t); for(int i = 1; i <= t; i++) { for(int j = n; j >= 0; j--) { if(j == n) //第i秒有n个人的概率等于第i－1秒有j－1个人的概率乘第i秒第j个人进电梯 //的概率加上第i - 1秒电梯就已经有n个人的概率 dp[i][j] = dp[i - 1][j - 1] * p + dp[i - 1][j]; else if(j != 0) //这里dp[i - 1][j]要乘(1 - p)表示第i秒排头不进电梯 dp[i][j] = dp[i - 1][j - 1] * p + dp[i - 1][j] * (1 - p); else //第i秒电梯里没人的概率为第i－1秒电梯里没人的概率乘(1 - p)表示 //第i秒时排头不进电梯 dp[i][j] = dp[i - 1][j] * (1 - p); } } for(int i = 1; i <= t; i++) ans += (dp[t][i] * i); printf("%.7f\n", ans); }