设f[i][j]为掉到f[i][j]时的概率然后分情况随便转移一下就好
主要是要手写分数比较麻烦
#include<iostream>
#include<cstdio>
using namespace std;
const int N=55;
int n,m;
char a[N][N];
long long gcd(long long a,long long b)
{
return !b?a:gcd(b,a%b);
}
struct fs
{
long long x,y;
fs(long long X=0,long long Y=1)
{
x=X,y=Y;
}
fs operator + (const fs &a) const
{
long long d=gcd(a.y,y),l=a.y/d*y;
fs b=fs(l/y*x+l/a.y*a.x,l);
long long g=gcd(b.x,b.y);
return fs(b.x/g,b.y/g);
}
fs operator * (const fs &a) const
{
fs b=fs(x*a.x,y*a.y);
long long g=gcd(b.x,b.y);
return fs(b.x/g,b.y/g);
}
}f[N][N];
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
for(int j=1;j<=i;j++)
{
a[i][j]=getchar();
while(a[i][j]!='*'&&a[i][j]!='.')
a[i][j]=getchar();
}
f[1][1]=fs(1,1);
for(int i=1;i<=n;i++)
for(int j=1;j<=i;j++)
{
if(a[i][j]=='*')
{
f[i+1][j]=f[i+1][j]+f[i][j]*fs(1,2);
f[i+1][j+1]=f[i+1][j+1]+f[i][j]*fs(1,2);
}
else
f[i+2][j+1]=f[i+2][j+1]+f[i][j];
}
printf("%lld/%lld\n",f[n+1][m+1].x,f[n+1][m+1].y);
return 0;
}
/*
5 2
*
*.
***
*.**
*****
*/