HDU 4987(概率dp）

Little Pony and Dice

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 489    Accepted Submission(s): 139

Problem Description

Twilight Sparkle was playing Ludo with her friends Rainbow Dash, Apple Jack and Flutter Shy. But she kept losing. Having returned to the castle, Twilight Sparkle became interested in the dice that were used in the game.

The dice has m faces: the first face of the dice contains a dot, the second one contains two dots, and so on, the m-th face contains m dots. Twilight Sparkle is sure that when the dice is tossed, each face appears with same probability. Also she knows that each toss is independent from others.

There is n + 1 cells in total, number from 0 to n, Twilight Sparkle race her token from 0 to n according to die rolls. The game end once she move to n or over it. Help Twilight Sparkle calculated the probability she end the game exactly at n.

 

Input

Input contains multiple test cases (less than 200). 
For each test case, only one line contains two integers m and n (1<=m, n<=10^9).

 

Output

For each case, output the corresponding result, rounded to 5 digits after the decimal point.

 

Sample Input

3 56 10

 

Sample Output

0.497940.28929

 

Source

BestCoder Round #7

 

题目链接：

题目描述：

大富翁游戏，多组输入m，n两个值，m表示筛子的最大点数，n表示格子数的最大编号（从0开始，总共n+1个各自）。当走到格子编号大于或等于n时则胜利。

求出当胜利时刚好到达n点时的概率。

解题思路：

这道题是概率dp，dp[i]表示刚好到达i时的概率。dp[0]=1,能到达dp[i-1]的点除了i-1-m外都能到达dp[i]。dp[i]=dp[i-1]+dp[i-1]*1.0/m,当i>m时dp[i]-=dp[i-m-1]*1.0/m。

当n的值越大时，dp[i-1]与dp[i]的值相当接近，所以可以用一个定值来表示。代码中用前后差值小于1e-13来表示非常小。

AC代码：

#include <iostream>#include <stdio.h>#include <cmath>#include <cstring>#include <algorithm>#define eps 1e-13#define maxn 703000//#define zero(x) (fabs(x)<eps?0:x)using namespace std;typedef long long LL;double zero(double x){    return ((fabs(x)<eps)? 0:x);}double dp[maxn];int main(){    LL m,n;    while(cin >> m >> n)    {        dp[0]=1.0;//初始位置概率为1        dp[1]=1.0/m;        int i;        for(i=2;i<=n;i++)        {            dp[i]=dp[i-1]+dp[i-1]*1.0/m;//转移公式            if(i>m)                dp[i]=dp[i]-dp[i-m-1]*1.0/m;//将i-m-1概率的贡献删去            if((zero(dp[i]-dp[i-1]))==0)//当i足够大师直接输出            {                printf("%.5lf\n",dp[i]);                break;            }        }        if(i>n)            printf("%.5lf\n",dp[n]);    }    return 0;}