poj 1189 钉子和小球

//这题有几个需要特别注意的地方
//不能算对应的格子的数量/总数，只能通过计算每层的概率得到
//一开始用计算总数，计得5/14和sample里的7/16对不上
//这里的层数有50层， 2^50次方，用unsigned long long
//在计算lcm时，要先除再乘，不然会溢出。
//在计算gcd时，传入的参数a,b要满足a>b
//因为这些原因各种错误RE，WA

#include<iostream>
using namespace std;

const int N = 60;

unsigned long long gcd(unsigned long long a, unsigned long long b)
{
	if (b == 0)
		return a;
	return gcd(b, a%b);
}

unsigned long long lcm(unsigned long long a, unsigned long long b)
{
	if (a > b)
		return a/gcd(a, b)*b;
	else 
		return b/gcd(b, a)*a;
}

struct fraction
{
	unsigned long long numerator;
	unsigned long long denominator; 
	fraction () {numerator = 0; denominator = 1;}
	fraction (unsigned long long numerator, unsigned long long denominator)
	{
		this->numerator = numerator;
		this->denominator = denominator;
	}

	friend fraction operator/ (const fraction& a, int n)
	{
		return fraction(a.numerator, a.denominator*n);
	}

	friend fraction operator+ (const fraction& a, const fraction& b)
	{
		fraction result;
		unsigned long long multiple, tmp;
		tmp = lcm(a.denominator, b.denominator);
		result.numerator = tmp/a.denominator*a.numerator + tmp/b.denominator*b.numerator;
		result.denominator = tmp;
		multiple = gcd(result.denominator, result.numerator);
		result.numerator /= multiple;
		result.denominator /= multiple;

		return result;
	}
};

char maze[N][N];
fraction dp[N][N];

int n, m;
int main()
{
	while (cin >> n >> m)
	{
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= i; j++)
			{
				cin >> maze[i][j];
				dp[i][j].denominator = 1;
				dp[i][j].numerator = 0;
			}

			for (int i = 1; i < N; i++)
			{
				maze[n+1][i] = '*';
				dp[n+1][i].denominator = 1;
				dp[n+1][i].numerator = 0;
			}

			for (int i = 1; i <= n+1; i++)
				for (int j = 1; j <= i; j++)
				{
					if (maze[i][j] == '*')
					{
						if (i == 1 && j == 1)
						{
							dp[i][j].numerator = 1;
							dp[i][j].denominator = 1;
						}					
						else if (i > 1 && j > 1 && j < i)
							dp[i][j] = dp[i-1][j-1]/2+dp[i-1][j]/2 + dp[i][j];
						else if (i > 1 && j == 1)
							dp[i][j] = dp[i-1][j]/2 + dp[i][j];
						else if (i > 1 && j == i)
							dp[i][j] = dp[i-1][j-1]/2 + dp[i][j];
					}
					else
					{
						dp[i][j].numerator = 0;
						dp[i][j].denominator = 1;
						int row = i+2;
						int count = 1;

						while (row <= n+1)
						{
							if (maze[row][j+count] == '*')
							{
								if (i == 1 && j == 1)
								{
									dp[row][j+count].numerator = 1;
									dp[row][j+count].denominator = 1;
								}
								else if (j > 1 && j < i)
									dp[row][j+count] = dp[i-1][j-1]/2 + dp[i-1][j]/2 + dp[row][j+count];
								else if (i > 1 && j == 1)
									dp[row][j+count] = dp[i-1][j]/2 + dp[row][j+count];
								else if (i > 1 && j == i)
									dp[row][j+count] = dp[i-1][j-1]/2 + dp[row][j+count];

								break;
							}
							else 
							{
								row += 2;
								count++;
							}
						}
					}
				}
		cout << dp[n+1][m+1].numerator << "/" << dp[n+1][m+1].denominator << endl;
	}
	return 0;
}