Codeforces Round #265 (Div. 2)E. Substitutes in Number(數學)，

Codeforces Round #265 (Div. 2)E. Substitutes in Number(數學)，

E. Substitutes in Numbertime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

Andrew and Eugene are playing a game. Initially, Andrew has string s, consisting of digits. Eugene sends Andrew multiple queries of type "di → ti", that means "replace all digits di in string s with substrings equal to ti". For example, if s = 123123, then query "2 → 00" transforms s to 10031003, and query "3 → " ("replace 3 by an empty string") transforms it to s = 1212. After all the queries Eugene asks Andrew to find the remainder after division of number with decimal representation equal to s by 1000000007 (109 + 7). When you represent s as a decimal number, please ignore the leading zeroes; also if s is an empty string, then it's assumed that the number equals to zero.

Andrew got tired of processing Eugene's requests manually and he asked you to write a program for that. Help him!

Input

The first line contains string s (1 ≤ |s| ≤ 105), consisting of digits — the string before processing all the requests.

The second line contains a single integer n (0 ≤ n ≤ 105) — the number of queries.

The next n lines contain the descriptions of the queries. The i-th query is described by string "di->ti", where di is exactly one digit (from 0 to 9), ti is a string consisting of digits (ti can be an empty string). The sum of lengths of ti for all queries doesn't exceed 105. The queries are written in the order in which they need to be performed.

Output

Print a single integer — remainder of division of the resulting number by 1000000007 (109 + 7).

Sample test(s)input

12312312->00

output

10031003

input

12312313->

output

1212

input

22222->00->7

output

777

input

10000000080

output

1

Note

Note that the leading zeroes are not removed from string s after the replacement (you can see it in the third sample).

題意:給你一個數字串和一些替換規則,要你輸出替換完之後模數的結果,可以用val[i]表示i表示的數,base[i]表示i所使用的進位.一個十進位串s可以表示成sum(10^(L-i)*s[i]),於是所有的變換都可轉化成乘法.代碼如下:

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int maxn=1e5+5;const int mod=1e9+7;string a,s[maxn];LL val[10],base[10];int f(char c){    return c-'0';}int main(){    ios_base::sync_with_stdio(false);    cin.tie(NULL);    int m;    cin>>a>>m;    for(int i=0;i<10;i++){        val[i]=i;        base[i]=10;    }    for(int i=1;i<=m;i++)cin>>s[i];    for(int i=m;i>=1;i--){        int L=s[i].size(),res=s[i][0]-'0';        LL v=0,b=1;        for(int j=L-1;j>2;j--){            v=(v+val[f(s[i][j])]*b)%mod;            b=(b*base[f(s[i][j])])%mod;        }        val[res]=v;        base[res]=b;        ///cout<<v<<' '<<b<<endl;    }    LL ans=0,b=1;    int L=a.size();    for(int i=L-1;i>=0;i--){        ans=(ans+b*val[f(a[i])])%mod;        b=(b*base[f(a[i])])%mod;    }    cout<<ans<<endl;    return 0;}