想状态和钉子的位置如何匹配想了半天。。。后来发现不是一样的吗$qwq$

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思路：当然是$DP$啦

提交：>5次（以为无故$RE$，实则是先乘后除爆了$long\space long$）

题解：

若有钉子，左右各乘$\frac{1}{2}$转移，否则，向下两层直接转移。

对于分数，分别维护分子和分母，然后加起来的时候，记着一定要写成

up[i][j]=up[i][j]*(b/G)+a*(dn[i][j]/G);
dn[i][j]=dn[i][j]*(b/G);

而非

up[i][j]=up[i][j]*b/G+a*dn[i][j]/G;
dn[i][j]=dn[i][j]*b/G;

（好吧也是我傻$qwq$）

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
#define ull unsigned long long
#define ll long long
#define R register ll 
#define pause (for(R i=1;i<=10000000000;++i))
#define In freopen("NOIPAK++.in","r",stdin)
#define Out freopen("out.out","w",stdout)
namespace Fread {
inline int g() {
    R ret=0,fix=1; register char ch; while(!isdigit(ch=getchar())) fix=ch=='-'?-1:fix;
    if(ch==EOF) return EOF; do ret=ret*10+(ch^48); while(isdigit(ch=getchar())); return ret*fix;
} inline bool isempty(const char& ch) {return (ch<=36||ch>=127);}
inline void gs(char* s) {
    register char ch; while(isempty(ch=getchar()));
    do *s++=ch; while(!isempty(ch=getchar()));
}
} using Fread::g; using Fread::gs;
namespace Luitaryi {
const int N=60;
int n,m;
ll up[N][N],dn[N][N];
bool w[N][N];
inline ll gcd(ll a,ll b){
    return b?gcd(b,a%b):a;
}
inline void add(int i,int j,ll a,ll b) {
    R G=gcd(dn[i][j],b);
    up[i][j]=up[i][j]*(b/G)+a*(dn[i][j]/G);
    dn[i][j]=dn[i][j]*(b/G);
    G=gcd(up[i][j],dn[i][j]); 
    if(G) up[i][j]/=G,dn[i][j]/=G;
}
inline void main() {
    n=g(),m=g()+1;
    for(R i=1;i<=n;++i) for(R j=1;j<=i;++j) { register char ch;
        while(ch=getchar(),ch!='*'&&ch!='.'); 
        w[i][j]=(ch=='*');
        up[i][j]=0,dn[i][j]=1;
    } for(R i=1;i<=n;++i) up[n+1][i]=0,dn[n+1][i]=1;
    up[1][1]=dn[1][1]=1;
    for(R i=1;i<=n;++i) for(R j=1;j<=i;++j) {
        R a=up[i][j],b=dn[i][j];
        if(w[i][j]) {
            if(a%2==0) a/=2; else b*=2;
            add(i+1,j,a,b),add(i+1,j+1,a,b);
        } else add(i+2,j+1,a,b);
    } printf("%lld/%lld",up[n+1][m],dn[n+1][m]);
    
}
}
signed main() {
    Luitaryi::main();    
}

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2019.07.17