AOJ-AHU-OJ-294 NBA Finals

[align=center]NBA Finals[/align]
[align=center]Time Limit: 1000 ms   Case Time Limit: 1000 ms   Memory Limit: 64 MB
[/align]
[align=center][/align]
Description
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.

Input
1st line: the game number (7<=n<=165) the winner played.

2nd line: the probability of Lakers winning a game p (0<p<1).

Output
1st line: The probability of Lakers winning the NBA Finals.

Round the result to 6 digits after the decimal point and keep trailing zeros.

Sample Input

Original

Transformed

7
0.4

Sample Output

Original

Transformed

0.228844

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思路：简单分析之后，知道了此题涉及两个问题：1.N局M胜(M=N/2+1)制 2.组合数

先解决组合数的问题。很明显，涉及阶乘。阶乘弄不好就会超时或者溢出。解决方法：打表。打表避免了递归。组合数的打表模板是这样的：

Com[0][0] = 1;
for(int i = 1; i < MAXN; i++)
for(int j = 0; j <= i; j++){
if(!j || j==i) Com[i][j] = 1;
else Com[i][j] = Com[i-1][j] + Com[i-1][j-1];
}
其次是排列组合的问题。对于N局M胜的概率问题，当时我推导了一遍公式。经验是：肯定是以胜局结尾。然后按一共打了几场分类讨论。公式如下：

i = 0 ——> i = game-win-1
ans += Com[game-i-1][win-1] × p^win × (1-p)^(game-win-i);代码如下：
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define MAXN 330
double Com[MAXN][MAXN];
double power(double a, int n)
{
double u = 1;
while(n--) u *= a;
return u;
}
int main(){
int win, game;//game表示是N局，win表示M胜
double p;
Com[0][0] = 1;
for(int i = 1; i < MAXN; i++)
for(int j = 0; j <= i; j++){
if(!j || j==i) Com[i][j] = 1;
else Com[i][j] = Com[i-1][j] + Com[i-1][j-1];
}
while(scanf("%d%lf", &win, &p)!=EOF){
double ans = 0;
game = win*2 - 1;
ans += power(p, win);//首先算出打了M场的时候，概率的值
for(int i = 0; i < game-win; i++)
ans += Com[game-i-1][win-1] * power(p, win) * power(1-p, game-win-i);
printf("%.6lf\n", ans);
}
return 0;
}