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~~
1 #include<iostream>
2 #include<cstring>
3 using namespace std;
4 int main()
5 {
6 double dp[201][201], p;
7 int i, j, n;
8 while (cin >> n >> p)
9 {
10 for (i = 1; i <= n; i++){
11 dp[i][0] = 0;
12 dp[0][i] = 1;
13 }
14 for (i = 1; i <= n; i++){
15 for (j = 1; j <= n; j++){
16 dp[i][j] = dp[i - 1][j] * p + dp[i][j - 1] * (1 - p);
17 }
18 }
19 cout << dp[n][n] << endl;
20 }
21 return 0;
22 }