/*
很明显这题就是一个乘法原理，我们求出每位上有可能的数字情况，
然后所有情况数相乘即为最终答案
我们可以考虑用Floyd的有向图的传递闭包
因为如果a可以转成b b可以转成c
那么a也可以转成c
注意这里数据过大一定要用高精度乘法
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

const int MAXN=505;
int ans[MAXN];
bool map[12][12];
int b[12];
int k;
string a;
int x,y;
int l=1;

void work(int x)//高精度乘法
{
    for(int i=0;i<l;i++)
        ans[i]*=x;
    for(int i=0;i<l;i++)
        if(ans[i]>=10)
        {
            ans[i+1]+=ans[i]/10;
            ans[i]%=10;
            l=max(l,i+2);
        }
}

int main()
{
    cin>>a;
    cin>>k;
    for(int i=1;i<=k;i++)//先建图
    {
        cin>>x>>y;
        map[x][y]=1;
    }
    for(int i=0;i<=9;i++)//自己到自己为1
        map[i][i]=1;
    for(int k=0;k<=9;k++)//Floyd求出所有能转换的关系
        for(int i=0;i<=9;i++)
            for(int j=1;j<=9;j++)
            if(map[i][k]&&map[k][j])
                map[i][j]=1;
    for(int i=0;i<=9;i++)//求出每个数字可以转换成多少个数字(注意包括自己)
    {
        int tot=0;
        for(int j=0;j<=9;j++)
            if(map[i][j])
                tot++;
        b[i]=tot;
    }
    ans[0]=1;//高精度初始化1
    for(int i=0;i<a.length();i++)
        work(b[a[i]-'0']);//乘法原理
    for(int i=l-1;i>=0;i--)
        cout<<ans[i];
    return 0;
}
     

对不起，看到大整数立马Python起来：

ss, n = map(str,input().split())
n = int(n)
test = [[0] * 10 for _ in range(10)]
acc = []
ans = 1

for i in range(n):
    x, y = map(int, input().split())
    test[x][y] = 1
for i in range(10):
    test[i][i] = 1
for k in range(10):
    for i in range(10):
        for j in range(10):
            if test[i][k] == 1 & test[k][j] == 1:
                test[i][j] = 1
for i in range(10):
    acc.append(0)
    for j in range(10):
        if test[i][j] == 1:
            acc[i] += 1
for i in range(len(ss)):
    ans *= acc[int(ss[i])]
print(ans);

/*
其实可以不用高精度
答案尾部有很多的0 把答案尾部的0的个数算出来 余下部分就可以long long
*/
#include<bits/stdc++.h>
using namespace std;

int f[12]={0};
int a[12][12]={0};
string m;
int n,i,j,k,x,y,t;
long long p;

int main(){
    cin >> m >> n;
    for (int i=1;i<=n;++i) {
        cin >> x >> y;
        a[x][y]=1;
    }
    for (int i=0;i<=9;++i) a[i][i]=1;
    for (int k=0;k<=9;++k)
        for (int i=0;i<=9;++i)
            for (int j=0;j<=9;++j)
                if (a[i][k] && a[k][j]) a[i][j]=1;
    for (int i=0;i<=9;++i) 
        for (int j=0;j<=9;++j)
            f[i]+=a[i][j];
    p=1; t=0;
    for (int i=0;i<m.length();++i){
        p*=f[m[i]-'0'];
        while (p%10==0) {
            t++;
            p/=10;
        }
    }
    cout << p;
    for (int i=1;i<=t;++i) cout << 0;
}

#include <iostream>                 //[2002普及组-C]产生数
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
int num[10], cnt;
int m[10][10];
bool Vis[10];
int number[1000], len;

void dfs(int x)
{
    for (int i = 0; i < 10; i++)
        if(m[x][i] && !Vis[i])
        {
            Vis[i] = true;
            cnt++;
            dfs(i);
        }
}

void mul(int number[], int k)
{
    int add = 0;
    for (int i = 0; i < len; i++)
    {
        int t = number[i] * k + add;
        number[i] = t % 10;
        add = t / 10;
    }
    while (add)
    {
        number[len++] = add % 10;
        add /= 10;
    }
}

int main()
{
    string n;
    int k, x, y;
    number[0] = 1, len = 1;
    cin >> n >> k;

    for (int i = 0; i < k; i++)
    {
        cin >> x >> y;
        m[x][y] = 1;
    }

    for (int i = 0; i < 10; i++)        //找出所有可达边
    {
        cnt = 0;
        memset(Vis, false, sizeof(Vis));
        Vis[i] = true;
        dfs(i);
        num[i] = cnt;
    }

    for (int i = 0; i < n.size(); i++)
        if(num[n[i] - '0'])
            mul(number, num[n[i] - '0'] + 1);

    for (int i = len - 1; i >= 0; i--)
        cout << number[i];
    cout << endl;

    system("pause");
    return 0;
}

#include<iostream>

#include<string.h>
#include<vector>
#include<map>
using namespace std;
map<char,vector<char> > m;
string s;
int b[10];
void dfs(char now){
    b[now-'0']=1;
    int len=m[now].size();
    for(int i=0;i<len;i++){
        if(!b[m[now][i]-'0']){
            dfs(m[now][i]);
        }
    }
}
int mul[100];
int main(){
    int k;
    cin>>s>>k;
    for(int i=1;i<=k;i++){
        char x,y;
        cin>>x>>y;
        m[x].push_back(y);
    }
    int l=s.size();
    mul[0]=1;
    for(int i=0;i<l;i++){
        memset(b,0,sizeof(b));
        dfs(s[i]);
        int sum=0;
        for(int i=0;i<=9;i++){
            if(b[i]){
                sum++;
            }
        }
        int x=0;
        for(int i=0;i<100;i++){
            mul[i]=mul[i]*sum+x;
            x=mul[i]/10;
            mul[i]%=10;
        }
    }
    int i=99;
    while(i>0&&!mul[i]){
        i--;
    }
    for(;i>=0;i--){
        cout<<mul[i];
    }
    return 0;
}

很水的题目，看清题意就好，直接把它转化为有向图，就是那个数字可以到那个数字，用folyd就好，因为时间就是O（100），folyd完后就直接乘法原理就可以了。

Var
        ch:char;
        i,j,k,num,n,x,y,l,t:longint;
        a,b,c:array[0..1000]of longint;
        f:array[0..9,0..9]of boolean;
Begin
        read(ch);
        while ch in ['0'..'9'] do
        begin
                inc(a[ord(ch)-48]);
                read(ch);
        end;
        readln(n);
        for i:=1 to n do
        begin
                readln(x,y);
                f[x,y]:=true;
        end;
        for i:=0 to 9 do
                f[i,i]:=true;

        for k:=0 to 9 do
                for i:=0 to 9 do
                        for j:=0 to 9 do
                                f[i,j]:=f[i,j] or (f[i,k] and (f[k,j]));

        for i:=0 to 9 do
        begin
                num:=0;
                for j:=0 to 9 do
                        if f[i,j] then
                                inc(num);
                c[i]:=num;
        end;
        b[1]:=1; l:=1;
        for k:=0 to 9 do
                for j:=1 to a[k] do
                begin
                        for i:=1 to l do
                        begin
                                b[i]:=b[i]*c[k]+t;
                                t:=b[i] div 10;
                                b[i]:=b[i] mod 10;
                        end;
                        if t>0 then
                        begin
                                inc(l);
                                b[l]:=t;
                                t:=0;
                        end;
                end;

        for i:=l downto 1 do
                write(b[i]);
        writeln;
        readln;
End.

Floyd+高精度秒杀
用Floyd求出每个数字可以变成多少种数字
然后乘起来
```c++
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 580 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 576 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 580 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 576 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 576 KiB, score = 10
Accepted, time = 0 ms, mem = 580 KiB, score = 50
代码
#include <iostream>
#include <string>
using namespace std;
string str;
int k,vis[10][10],f[10],num[101];
inline void floyd() {
for (int k = 0;k <= 9;k++)
for (int i = 0;i <= 9;i++)
for (int j = 0;j <= 9;j++) vis[i][j] = vis[i][j] || (vis[i][k] && vis[k][j]);
}
int main (){
ios::sync_with_stdio(false);
cin >> str >> k;
while (k--) {
int a,b;
cin >> a >> b;
vis[a][b] = true;
}
for (int i = 0;i <= 9;i++) vis[i][i] = true;
floyd();
for (int i = 0;i <= 9;i++)
for (int j = 0;j <= 9;j++)
if (vis[i][j]) f[i]++;
int len = 2; num[1] = 1;
for (int i = 0;i < (int)str.length();i++) {
for (int j = 1;j <= 100;j++)
num[j] *= f[str[i]-'0'];
for (int j = 1;j <= 100;j++)
if (num[j] >= 10) {
num[j+1] += num[j]/10;
num[j] %= 10;
}
while (num[len]) len++;
}
for (int i = len-1;i >= 1;i--) cout << num[i];
return 0;
}
```

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;

const int Maxn=31;

char s[Maxn];
int k,len,len1,tot[Maxn],ans[Maxn],dis[Maxn][Maxn];

void mul(int c)
{
int x=0;
for(int i=1;i<=len1;i++){
ans[i]=ans[i]*c+x;
x=ans[i]/10;
ans[i]%=10;
}
while(x>0){ans[++len1]=x%10;x/=10;}
}

int main()
{
ans[1]=1;
len1=1;
for(int i=0;i<=9;i++)tot[i]=1;
cin>>s+1;
len=strlen(s+1);
scanf("%d",&k);
for(int i=1;i<=k;i++){
int x,y;
scanf("%d%d",&x,&y);
if(!dis[x][y])tot[x]++;
dis[x][y]=1;//建图
}

for(int k=0;k<=9;k++){//听起来很牛B的传递闭包(floyd实现)
for(int i=0;i<=9;i++)
if(dis[i][k]){
for(int j=0;j<=9;j++)
if(dis[i][k]&&dis[k][j]){
if(i!=j&&!dis[i][j])tot[i]++;
dis[i][j]=1;
}
}
}

for(int i=1;i<=len;i++)
if(tot[s[i]-48]!=1){
mul(tot[s[i]-48]);
}

for(int i=len1;i>=1;i--)printf("%d",ans[i]);
}
高精乘低精即可 加floyd传递闭包可秒杀

这个题读题有点障碍，应该理解成，
变换后的数字依旧可以变换
例如：
原式 123 规则 3-7 7-4
就可以变成127再变成124
基本思路：
Floyd（有向图）类似的方式算出0-9可以变成的数字的个数 高精度chg[] ：
floyd可以算出最短路 但这里只需要知道哪些点相连 只需map[i][j]>0 (初始为0)
如果3点与7点相连 （有向） 就说明3可以变成7 执行chg[3]++
自己也可以连向自己

int count[x]代表x在原式出现的次数
高精度 sum=1
for(i=1~9) 高精度乘法 sum*=pow(chg[i],count[i])

sum为所求

附上代码

#include <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#define maxn 50 

void multx(int a[],int b[],int c[]){
    memset(c,0,sizeof(int)*maxn);
    for(int i=1;i<=a[0];i++)
    for(int j=1;j<=b[0];j++){
        c[i+j-1]+=a[i]*b[j];
        c[i+j]+=c[i+j-1]/10000;
        c[i+j-1]%=10000;
    }
    int len=a[0]+b[0];
    while(c[len]==0&&len>1)
        len--;
    c[0]=len;
}

void output(int a[]){
    printf("%d",a[a[0]]);
    for(int i=a[0]-1;i>=1;i--)
        printf("%04d",a[i]);
}

int main(){
//  freopen("in.txt","r",stdin);
    int count[10]={0},p,map[10][10]={0};
    int chg[10][maxn]={0};
    for(int i=0;i<=9;i++)
        chg[i][0]=1;
        
    int sum[maxn]={1,1};
    char a[100];
    std::cin>>a>>p;
    for(int i=1;i<=p;i++){
        int z,j;
        scanf("%d%d",&z,&j);
        map[z][j]=1;
    }
    for(int k=0;k<=9;k++)
    for(int i=0;i<=9;i++)
    for(int j=0;j<=9;j++)
        if(map[i][k]>0&&map[k][j]>0)
            map[i][j]=1;
            
    for(int i=0;i<strlen(a);i++)
        count[a[i]-'0']++;
        
    for(int i=0;i<=9;i++)
    for(int j=0;j<=9;j++)
        if(i==j||map[i][j]>0)
            chg[i][1]++;    
                
    int temp[maxn],t[maxn],t2[maxn];        
    for(int i=0;i<=9;i++)
        if(count[i]>0){
            memcpy(temp,chg[i],sizeof(temp));
            memcpy(t2,temp,sizeof(temp));
            
            for(int x=1;x<=count[i]-1;x++){
                multx(t2,temp,t);
                memcpy(temp,t,sizeof(temp));
            }
            
            multx(sum,temp,t);
            memcpy(sum,t,sizeof(t));
        }
        
    output(sum);
    return 0;
}

var
ch:char;
s:string;
f:array [0..10,0..10] of longint;
a,b,c,d:array [0..1005] of longint;
n,i,j,k,l,x,y:longint;
begin
read(ch);
while (ch<>' ') do begin inc(a[ord(ch)-48]); read(ch); end;
read(n); fillchar(f,sizeof(f),10);
for i:=1 to n do begin read(x,y); f[x,y]:=0; end;
for i:=0 to 9 do f[i,i]:=0;
for k:=0 to 9 do
for i:=0 to 9 do
for j:=0 to 9 do
if (f[i,j]>f[i,k]+f[k,j]) then f[i,j]:=f[i,k]+f[k,j];
for i:=0 to 9 do
for j:=0 to 9 do if f[i,j]=0 then inc(c[i]);
d[1]:=1; l:=1;
for i:=0 to 9 do
for j:=1 to a[i] do
begin
for k:=1 to l do d[k]:=d[k]*c[i];
for k:=1 to l do begin inc(d[k+1],d[k] div 10); d[k]:=d[k] mod 10; end;
if (d[l+1]<>0) then inc(l);
end;
for i:=l downto 1 do write(d[i]);
end.

这个n<10的30次方是什么鬼？O(N)都过不了啊

条件是什么意思？急！！！！

记录信息
评测状态 Accepted
题目 P1129 产生数
递交时间 2015-10-06 22:19:31
代码语言 C++
评测机 VijosEx
消耗时间 2 ms
消耗内存 532 KiB
评测时间 2015-10-06 22:19:32
评测结果
编译成功

foo.cpp: In function 'int main()':
foo.cpp:40:10: warning: statement has no effect [-Wunused-value]
for(cnt;cnt<=maxn||in;cnt++)
^
测试数据 #0: Accepted, time = 0 ms, mem = 524 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 528 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 528 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 532 KiB, score = 10
测试数据 #4: Accepted, time = 2 ms, mem = 524 KiB, score = 10
Accepted, time = 2 ms, mem = 532 KiB, score = 50
代码
#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
int num[10][10];
int nk[10];
int ans[123];
char s[123];
int main()
{
int kn;
scanf("%s",s+1);
scanf("%d",&kn);

s[0]='1';
memset(num,0,sizeof(num));
int len=strlen(s)-1;
int a,b;
for(int i=1;i<=kn;i++)
{
scanf("%d%d",&a,&b);
num[a][b]=1;
}
for(int i=0;i<=9;i++)num[i][i]=1;
//floyd算法
for(int k=0;k<=9;k++)
for(int i=0;i<=9;i++)
for(int j=0;j<=9;j++)
if(num[i][k]==1&&num[k][j]==1)
num[i][j]=1;
for(int i=0;i<=9;i++)
for(int j=0;j<=9;j++)
nk[i]+=num[i][j];
//高精度
int maxn=1;
ans[1]=1;
for(int pos=1;pos<=len;pos++)
{
int nc=nk[s[pos]-'0'];
int in=0,cnt=1;
for(cnt;cnt<=maxn||in;cnt++)
{

ans[cnt]=nc*ans[cnt]+in;
in=ans[cnt]/10;
ans[cnt]%=10;
}
if(cnt-1>maxn)maxn=cnt-1;
}

for(int i=maxn;i>=1;i--)printf("%d",ans[i]);

}

这题目真是困扰了我好久......自己去网上了解“传递闭包”这个概念然后做了很多遍，都是60分，现在终于AC
关键就是传递闭包时i,j,k的顺序，应该把中间量j放在循环最外层，这样的传递才不会因输入顺序不同造成遗漏
开始还以为是精度问题，所以最后写了高精度

#include<stdio.h>
#include<string.h>
int main( )
{
int gjd[100]={0};
char g[31];
int a,i,k[11]={0},x,y,n,fy[11][11],j,l,pot=1,zc,xx;
scanf("%s",&g);
scanf("%d",&a);
n=strlen(g);

for(i=0;i<=9;i++)
for(j=0;j<=9;j++)
fy[i][j]=0;

for(i=1;i<=a;i++)
{
scanf("%d %d",&x,&y);
if(y!=0) fy[x][y]=1;
}

for(i=0;i<=9;i++)
fy[i][i]=1;

for(j=0;j<=9;j++)
for(i=0;i<=9;i++)
for(l=0;l<=9;l++)

if(fy[i][j]==fy[j][l] && fy[i][j]==1) fy[i][l]=1;

for(i=0;i<=9;i++)
for(j=0;j<=9;j++)
if(fy[i][j]==1) k[i]=k[i]+1;

gjd[1]=1;

for(i=0;i<=n-1;i++)
{
for(j=1;j<=pot;j++)
gjd[j]=gjd[j]*(k[g[i]-48]);
zc=pot;
for(j=1;j<=zc;j++)
{
if(gjd[j]>=10) {xx=gjd[j]/10;
gjd[j]=gjd[j]-10*xx;
gjd[j+1]=gjd[j+1]+xx;
if(j==zc) pot++;}
}

}

for(i=pot;i>=1;i--)

printf("%d",gjd[i]);

return 0;
}

program Project1;

var
v, u, x, y: array[0..9] of longint;
a: array[0..9, 0..9] of longint;
b: array[1..1000] of longint;
sum: array[1..10000] of longint;
i, t1, k, t, Result, j, h: longint;
c: char;
w: array[0..9, 0..9] of boolean;

procedure cz;
var

i, j, k1: longint;
begin

for i := 0 to 9 do
w[i, i] := True;
for k1 := 0 to 9 do
for i := 0 to 9 do
for j := 0 to 9 do

begin
if (k1 = i) or (i = j) or (i = k1) then
h := 1
else
if (w[i, j] = False) and (w[i, k1] = True) and (w[k1, j] = True) and
(a[i, j] < a[i, k1] + a[k1, j]) then
begin
w[i, j] := True;
a[i, j] := a[i, k1] + a[k1, j];
end;
end;
for i := 0 to 9 do
for j := 0 to 9 do
if w[i, j] then
Inc(v[i]);

end;

procedure fj;
var
c: array[1..1000] of longint;
x: longint;
begin
fillchar(b, sizeof(b), 0);
fillchar(c, sizeof(c), 0);
x := v[i];
t := 0;
while x <> 0 do
begin
Inc(t);
c[t] := x mod 10;
x := x div 10;
end;
for j := 1 to t do
b[j] := c[j];
end;

procedure gj;
var
j, k: longint;
su: array[1..10000] of longint;
begin
fillchar(su, sizeof(su), 0);
for j := 1 to t1 do
su[j] := sum[j];
fillchar(sum, sizeof(sum), 0);
for j := 1 to t1 do
for k := 1 to t do
begin
sum[j + k - 1] := su[j] * b[k] + sum[j + k - 1];
if sum[j + k - 1] > 9 then
begin
sum[j + k] := sum[j + k - 1] div 10;
sum[j + k - 1] := sum[j + k - 1] mod 10;
if t1 < j + k then
t1 := j + k;
end;
if t1 < j + k - 1 then
t1 := j + k - 1;
end;
end;

begin
Read(c);
fillchar(w, sizeof(w), False);
while c <> ' ' do
begin
val(c, i, Result);
Inc(u[i]);
Read(c);
end;
Read(k);
readln;
for i := 1 to k do
begin
readln(x[i], y[i]);
a[x[i], y[i]] := 1;
w[x[i], y[i]] := True;
end;
cz;
sum[1] := 1;
t1 := 1;
h := 0;
for i := 0 to 9 do
begin
if (u[i] <> 0) and (v[i] <> 0) then
begin
fj;
for j := 1 to u[i] do
gj;
end;
end;
for i := t1 downto 1 do
Write(sum[i]);
end.

对于每一位，统计出他能变换成哪些数字（包括他本身），之后每一位都有ai种选择，所以根据乘法原理，将每一位的选择数相乘，但是结果得高精。至于统计选择的数量，用FLOYD即可。
下面贴代码，我只是个蒟蒻，写得麻烦的话求不喷。
#include<iostream>
#include<cstdio>
using namespace std;
int kps[30]={0};
void p(int x)
{
for(int i=1;i<=kps[0];i++)kps[i]=kps[i]*x;for(int i=1;i<=kps[0];i++){kps[i+1]=kps[i+1]+(kps[i]/10);kps[i]=kps[i]%10;}
for(kps[0]++;kps[kps[0]]!=0;kps[0]++){kps[kps[0]+1]=kps[kps[0]+1]+(kps[kps[0]]/10);kps[kps[0]]=kps[kps[0]]%10;}kps[0]--;
}
int main()
{
bool map[11][11]={false};int n,k;
char calc[100];int ans=0;
for(scanf("%c",&calc[0]);calc[ans]!=' ';ans++){scanf("%c",&calc[ans+1]);}
scanf("%d",&k);
for(int i=1;i<=k;i++){int o,l;scanf("%d%d",&o,&l);map[o][l]=true;}
for(int i=0;i<=9;i++)map[i][i]=true;for(int k=0;k<=9;k++)for(int i=0;i<=9;i++)for(int j=0;j<=9;j++)if(map[i][k]==true)if(map[k][j]==true)map[i][j]=true;
int f[10]={0};for(int i=0;i<=9;i++)for(int j=0;j<=9;j++)if(map[i][j]==true)f[i]++;kps[0]=1;kps[1]=1;for(int i=0;i<=ans-1;i++){p(f[(calc[i]-'0')]);n=n/10;}
for(int i=kps[0];i>0;i--)printf("%d",kps[i]);return 0;
}

var
s:string;
j,i,k,p,x,y,n:longint;
f:array[0..9,0..9] of boolean;
ans:real;
begin
readln(s);
p:=pos(' ',s);
val(copy(s,p+1,100),n);
delete(s,p,100);
for i:=1 to n do
begin
readln(x,y);
f[x,y]:=true;
end;
for k:=0 to 9 do
for i:=0 to 9 do
for j:=0 to 9 do
if (i<>j) and (i<>k) and (k<>j) then f[i,j]:=f[i,j] or f[i,k] and f[k,j];
ans:=1;
for i:=1 to length(s) do
begin
j:=ord(s[i])-48;
p:=1;
for k:=0 to 9 do
if f[j,k] then inc(p);
ans:=ans*p;
end;
writeln(ans:0:0);
end.
膜拜LGS大牛。
本来要老老实实高精。
他的建议太强了！
只是担心比赛时万一出了精度问题。。。。。

其实不用写高精的，把ANS开成real，然后取整就AC了，一遍秒杀！；FOLYED;
L.G.S-庆祝国庆节

why? Who can help me?
VijosEx via JudgeDaemon2/13.7.4.0 via libjudge

编译成功

测试数据 #0: WrongAnswer, time = 15 ms, mem = 732 KiB, score = 0

测试数据 #1: WrongAnswer, time = 15 ms, mem = 728 KiB, score = 0

测试数据 #2: WrongAnswer, time = 15 ms, mem = 732 KiB, score = 0

测试数据 #3: WrongAnswer, time = 15 ms, mem = 732 KiB, score = 0

测试数据 #4: WrongAnswer, time = 0 ms, mem = 728 KiB, score = 0

WrongAnswer, time = 60 ms, mem = 732 KiB, score = 0

var
i,j,k,e,y,x,p,f,n:integer;
ch:char;
a,s,c:array[1..50] of integer;
b:array[1..15] of integer;
procedure ready;
begin
n:=0;
y:=0;
read(ch);
while ch<>' ' do
begin
inc(n);
a[n]:=ord(ch)-ord('0');
read(ch);
end;
readln(k);
for x:=1 to k do readln(b[x]);
for j:=1 to n do
c[j]:=1;
for e:=1 to n do
for j:=1 to k do
if b[j]=a[e] then inc(c[e]);
end;
procedure gao;
var i,t,max,ma:integer;
begin
for i:=1 to 50 do s[i]:=0;
s[1]:=1;
i:=1; max:=1;
while i<=n do
begin
j:=1;
while j<=max do
begin
ma:=max;
s[j]:=s[j]*c[i];
p:=j;
while s[p]>=10 do
begin
t:=s[p];
s[p]:=t mod 10;
s[p+1]:=s[p+1]+(t div 10);
inc(p);
end;
if p>ma then ma:=p;
inc(j);
end;
if ma>max then max:=ma;
inc(i);
end;
end;
procedure print;
begin
for f:=1 to p do write(s[f]);
end;
begin
ready;
gao;
print;
end.

算法：DFS + 高精度乘法 + 乘法原理