codeforces 464C. Substitutes in Number

题目链接

C. Substitutes in Number

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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

123123
1
2->00

output

10031003

input

123123
1
3->

output

1212

input

222
2
2->0
0->7

output

777

input

1000000008
0

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).

给一个字符串， m个操作， 每次操作， 将其中的一个数变为一个字符串， 问最后的字符串%1e9+7的结果。

 #include <iostream>
 #include <vector>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <map>
 #include <set>
 #include <string>
 #include <queue>
 #include <stack>
 #include <bitset>
 using namespace std;
 #define pb(x) push_back(x)
 #define ll long long
 #define mk(x, y) make_pair(x, y)
 #define lson l, m, rt<<1
 #define mem(a) memset(a, 0, sizeof(a))
 #define rson m+1, r, rt<<1|1
 #define mem1(a) memset(a, -1, sizeof(a))
 #define mem2(a) memset(a, 0x3f, sizeof(a))
 #define rep(i, n, a) for(int i = a; i<n; i++)
 #define fi first
 #define se second
 typedef pair<int, int> pll;
 const double PI = acos(-1.0);
 const double eps = 1e-;
 const int mod = 1e9+;
 const int inf = ;
 const int dir[][] = { {-, }, {, }, {, -}, {, } };
 const int maxn = 1e5+;
 ll val[], Pow[];
 int d[maxn];
 char c[maxn];
 string s[maxn], str;
 int main()
 {
     cin>>str;
     int n;
     scanf("%d", &n);
     for(int i = ; i<n; i++) {
         scanf("%d->", &d[i]);
         gets(c);
         s[i] = c;
     }
     for(int i = ; i<; i++) {
         val[i] = i;
         Pow[i] = ;
     }
     for(int i = n-; i>=; i--) {　　　　　　//这里倒着做...
         ll p = , v = ;
         int len = s[i].size();
         for(int j = ; j<len; j++) {
             int tmp = s[i][j]-'';
             p = p*Pow[tmp]%mod;
             v = v*Pow[tmp]%mod;
             v = (v+val[tmp])%mod;
         }
         val[d[i]] = v;
         Pow[d[i]] = p;
     }
     int len = str.size();
     ll ans = ;
     for(int i = ; i<len; i++) {
         ans = (ans*Pow[str[i]-'']%mod+val[str[i]-''])%mod;
     }
     printf("%d\n", ans);
     return ;
 }