[Swust OJ 649]--NBA Finals(dp,后台略(hen)坑)

题目链接：http://acm.swust.edu.cn/problem/649/

Time limit(ms): 1000　　　　Memory limit(kb): 65535

Consider two teams, Lakers and Celtics, playing a series of 
NBA Finals until one of the teams wins n games. Assume that the probability 
of Lakers winning a game is the same for each game and equal to p and 
the probability of Lakers losing a game is q = 1-p. Hence, there are no 
ties.Please find the probability of Lakers winning the NBA Finals if the 
probability of it winning a game is p.

Description

first line input the n-games (7<=n<=165)of NBA Finals 
second line input the probability of Lakers winning a game p (0< p < 1)

Input

the probability of Lakers winning the NBA Finals

Output

1 2

7 0.4

Sample Input

1

0.289792

Sample Output

 

题目大意：假设湖人和凯尔特人在打NBA总决赛，直到一支队伍赢下 n 场比赛，那只队伍就获得总冠军，

　　　　　假定湖人赢得一场比赛的概率是 p，即输掉比赛的概率为 1-p，每场比赛不可能以平局收场，问湖人赢得这个系列赛的概率

 

解题思路：这就是一个数学题(貌似)高中的概率题，明显一个dp问题，P[i][j]的含义是：当A队还有 i 场比赛需要赢，

      　　　才能夺得冠军，B队还有 j 场比赛需要赢，才能夺得冠军时，A队获得冠军的概率，

　　　　　边界 P[i][0]=0 (1<=i<=n)（B队已经夺冠了），P[0][i]=1 (1<=i<=n)（A队已经夺冠了），

　　　　　要求的输出即是 P[n][n]。

 

最后友情提示一下：学校OJ(swust oj)太坑,居然输出dp[n-3][n-3]~~受不了~~

#include<iostream>
#include<cstring> 
using namespace std;
int main()
{
    double dp[201][201], p;
    int i, j, n;
    while (cin >> n >> p)
    {
        for (i = 1; i <= n; i++){
            dp[i][0] = 0;
            dp[0][i] = 1;
        }
        for (i = 1; i <= n; i++){
            for (j = 1; j <= n; j++){
                dp[i][j] = dp[i - 1][j] * p + dp[i][j - 1] * (1 - p);
            }
        }
        cout << dp[n][n] << endl;
    }
    return 0;
}