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

http://codeforces.com/contest/465/problem/E

给定一个字符串，以及n个变换操作，将一个数字变成一个字符串，可能为空串，然后最后将字符串当成一个数，取模1e9+7。

逆向操作，维护每次替换后产生的数值和长度

替换P - > d_1d_2……d_n后

新的P的长度Len[ d_1 ] +……+ Len [ d_n ]

新的P值是Val[ d_n ] + 10 ^（Len [ d_n ]）* Val [ d_（n-1）] + 10 ^（Len [ d_n ] + [ d_ len（n-1）]）* Val [ d_（n-2）] +…10 ^（Len [ d_n ] + [ d_ len（n-1）] +……+ Len [ d_2 ]）* Val [ d_1 ]取模。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <cassert>
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define clr0(x) memset(x,0,sizeof(x))
typedef long long LL;
const int maxn = 100005,modo = 1000000007;
char s[maxn],_q[maxn];
int d[maxn];
string q[maxn];
int n,rep[10],pow[10];
int main() {
scanf("%s%d",s,&n);
for(int i = 0;i < n;++i){
//scanf("%d->%s",&d[i],_q);
scanf("%s",_q);
d[i] = _q[0] - '0';
q[i] = _q+3;
}
for(int i = 0;i < 10;++i)
rep[i] = i,pow[i] = 10;
for(int i = n-1;i >= 0;--i){
LL r = 0,p = 1LL;
for(int j = 0;j < q[i].size();++j){
int num = q[i][j] - '0';
p = (p*pow[num])%modo;
r = (r*pow[num] + rep[num])%modo;
}
rep[d[i]] = r;
pow[d[i]] = p;
}
n = strlen(s);
LL ans = 0;
for(int i = 0;i < n;++i){
int num = s[i]-'0';
ans = (ans*pow[num]+rep[num])%modo;
}
printf("%I64d\n",ans);
return 0;
}