codeforces 518D. Ilya and Escalator 概率dp

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

题意: 有n个人站队上电梯, 每一秒站在队首的人有p的概率上电梯, 或者(1-p)的概率不上电梯, 如果上电梯, 只能上一个人, 如果不上电梯, 那么后面的人也不能上电梯, 求t秒之后电梯上人数的期望.

分析: 可以令dp[i][j]表示时间为i的时候有j个人的概率, 那么有状态转移为dp[i+1][j+1] = p*(dp[i][j]), 当有n个人的时候要特殊转移一下.

#include<bitset>
#include<map>
#include<vector>
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<stack>
#include<queue>
#include<set>
#define inf 0x3f3f3f3f
#define mem(a,x) memset(a,x,sizeof(a))
#define F first
#define S second
using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

inline int in()
{
    int res=0;char c;int f=1;
    while((c=getchar())<'0' || c>'9')if(c=='-')f=-1;
    while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
    return res*f;
}
const int N=100010,MOD=1e9+7;
double dp[2345][2345];

int main()
{
    int n,t;
    double p;
    cin>>n>>p>>t;
    dp[0][0]=1.0;
    for(int i=0;i<t;i++){
        for(int j=0;j<=n;j++){
            if(j==n) dp[i+1][j] += dp[i][j];
            else{
                dp[i+1][j+1] += p*dp[i][j];
                dp[i+1][j] += (1.0-p)*dp[i][j];
            }
        }
    }
    double ans=0;
    for(int i=0;i<=n;i++) ans += dp[t][i]*i;
    printf("%.9lf\n",ans);
    return 0;
}

codeforces 518D. Ilya and Escalator 概率dp

猜你喜欢