化环为链，FFT做字符串匹配即可。
Code:

#include<bits/stdc++.h>
using namespace std;

#define ll long long int
//Header

namespace fft
{
    const double pi=acos(-1);
    int limit=1,l=0;
    struct cp{//complex
        double r,i;
        cp(double real=0,double img=0){r=real;i=img;}
        void set(double real=0,double img=0){r=real;i=img;}
        cp operator+(const cp& a){return cp(r+a.r,i+a.i);}
        cp operator-(const cp& a){return cp(r-a.r,i-a.i);}
        cp operator*(const cp& a){return cp(r*a.r-i*a.i,r*a.i+i*a.r);}
    };
    
    void FFT(vector<cp> &a,int type)
    {
        vector<int>r(limit+1);
        //int r[limit+1];r[0]=0;
        for(int i=0;i<limit;i++)r[i]=(r[i>>1]>>1 | (i&1)<<(l-1));
        for(int i=0;i<limit;i++)if(i<r[i])swap(a[i],a[r[i]]);
        cp Wn,w;
        for(int mid=1;mid<limit;mid<<=1)//不断取出中点
        {
            Wn.set(cos(pi/mid),type*sin(pi/mid));//因 mid只有一半 所以就不用乘2了。
            int size=mid<<1;
            for(int i=0;i<limit;i+=size)
            {
                w.set(1,0);
                int j=i+mid;//计算前一半
                for(int k=i;k<j;k++,w=w*Wn)
                {
                    cp x=a[k],y=w*a[k+mid];
                    a[k]=x+y;
                    a[k+mid]=x-y;
                }
            }
        }
    }

    void mul_fft(vector<cp> &a, vector<cp> &b)
    {
        int siz=a.size()+b.size();
        limit=1;l=0;
        while(limit<siz)limit*=2,l++;
        a.resize(limit);b.resize(limit);
        for(int i=0;i<limit;i++)a[i].i=b[i].r;
        FFT(a,1);for(int i=0;i<limit;i++)a[i]=a[i]*a[i];FFT(a,-1);//三次变两次优化 
        for(int i=0;i<=siz;i++) a[i].r=a[i].i/limit/2;
    }
}

string s1,s2;
double ans[1000005];
int s[3][3];
void solve()
{
    for(int i=0;i<=2;i++)for(int j=0;j<=2;j++)cin>>s[i][j];
    string s1,s2;cin>>s1>>s2;
    s1+=s1;
    char ch[]={'R','G','B'};
    reverse(s2.begin(),s2.end());
    int n = s1.length(),m=s2.length();
    for(int i=0;i<=n+m;i++)ans[i]=0;
    for(int i=0;i<3;i++)
    {
        vector<fft::cp> a(n+m+1),b(m+1);
        for(int j=0;j<n;j++) 
        {
            if(s1[j]==ch[i]) a[j]=fft::cp(1,0); else a[j]=fft::cp(0,0);
        } 
        for(int j=0;j<m;j++) 
        {
            if(s2[j]=='R')b[j]=fft::cp(s[i][0],0);
            if(s2[j]=='G')b[j]=fft::cp(s[i][1],0);
            if(s2[j]=='B')b[j]=fft::cp(s[i][2],0);
        } 
        fft::mul_fft(a,b);
        for(int j=m;j<=n-1;j++) ans[j]+=a[j].r;
    }
    ll res=0;
    for(int i=m;i<=n-1;i++)res=max(res,(ll)(ans[i]+0.5));
    cout<<res<<endl;
}
int main()
{
    solve();
    return 0;
}

#include<bits/stdc++.h>
using namespace std;

#define ll long long int
//Header

namespace fft
{
const double pi=acos(-1);
int limit=1,l=0;
struct cp{//complex
double r,i;
cp(double real=0,double img=0){r=real;i=img;}
void set(double real=0,double img=0){r=real;i=img;}
cp operator+(const cp& a){return cp(r+a.r,i+a.i);}
cp operator-(const cp& a){return cp(r-a.r,i-a.i);}
cp operator*(const cp& a){return cp(r*a.r-i*a.i,r*a.i+i*a.r);}
};

void FFT(vector<cp> &a,int type)
{
vector<int>r(limit+1);
//int r[limit+1];r[0]=0;
for(int i=0;i<limit;i++)r[i]=(r[i>>1]>>1 | (i&1)<<(l-1));
for(int i=0;i<limit;i++)if(i<r[i])swap(a[i],a[r[i]]);
cp Wn,w;
for(int mid=1;mid<limit;mid<<=1)//不断取出中点
{
Wn.set(cos(pi/mid),type*sin(pi/mid));//因 mid只有一半 所以就不用乘2了。
int size=mid<<1;
for(int i=0;i<limit;i+=size)
{
w.set(1,0);
int j=i+mid;//计算前一半
for(int k=i;k<j;k++,w=w*Wn)
{
cp x=a[k],y=w*a[k+mid];
a[k]=x+y;
a[k+mid]=x-y;
}
}
}
}

void mul_fft(vector<cp> &a, vector<cp> &b)
{
int siz=a.size()+b.size();
limit=1;l=0;
while(limit<siz)limit*=2,l++;
a.resize(limit);b.resize(limit);
for(int i=0;i<limit;i++)a[i].i=b[i].r;
FFT(a,1);for(int i=0;i<limit;i++)a[i]=a[i]*a[i];FFT(a,-1);//三次变两次优化
for(int i=0;i<=siz;i++) a[i].r=a[i].i/limit/2;
}
}

string s1,s2;
double ans[1000005];
int s[3][3];
void solve()
{
for(int i=0;i<=2;i++)for(int j=0;j<=2;j++)cin>>s[i][j];
string s1,s2;cin>>s1>>s2;
s1+=s1;
char ch[]={'R','G','B'};
reverse(s2.begin(),s2.end());
int n = s1.length(),m=s2.length();
for(int i=0;i<=n+m;i++)ans[i]=0;
for(int i=0;i<3;i++)
{
vector<fft::cp> a(n+m+1),b(m+1);
for(int j=0;j<n;j++)
{
if(s1[j]==ch[i]) a[j]=fft::cp(1,0); else a[j]=fft::cp(0,0);
}
for(int j=0;j<m;j++)
{
if(s2[j]=='R')b[j]=fft::cp(s[i][0],0);
if(s2[j]=='G')b[j]=fft::cp(s[i][1],0);
if(s2[j]=='B')b[j]=fft::cp(s[i][2],0);
}
fft::mul_fft(a,b);
for(int j=m;j<=n-1;j++) ans[j]+=a[j].r;
}
ll res=0;
for(int i=m;i<=n-1;i++)res=max(res,(ll)(ans[i]+0.5));
cout<<res<<endl;
}
int main()
{
solve();
return 0;
}