CodeForces 464 B Restore Cube

Restore Cube

题解：

x->yyy

其实就是把x代替成yyy这个值。

如果不好理解的话， 可以试想一下， 刚开始的话 0->0, 1->1, 2->2,...,9->9.

现在有一条指令 1->23

那么就是就是0->0, 1->23， 2->2,...,9->9.

现在又有一条指令2->45

那么就相当于0->0, 1->453, 2->45,...,9->9.

就相当于我们求出0~9每个数字分别代表的是什么东西。

然后我们倒着DP给定的指令。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+;
const int N = 1e5 + ;
char s[N];
string str[N];
LL dp[], base[];
void solve(string & sstr){
    int m = sstr[] - '';
    int len = sstr.size();
    LL tmp_base = ; LL tmp_yu = ;
    for(int i = ; i < len; ++i){
        int id = sstr[i] - '';
        tmp_base = (tmp_base * base[id]) % mod;
        tmp_yu = (tmp_yu * base[id] + dp[id]) % mod;
    }
    dp[m] = tmp_yu, base[m] = tmp_base;
//    cout << m << ' ' << dp[m] << ' '  << base[m] << endl;
}
int main(){
    scanf("%s", s+);
    int n;
    scanf("%d", &n);
    for(int i = ; i < ; ++i)
        dp[i] = i, base[i] = ;
    for(int i = ; i <= n; ++i){
        cin >> str[i];
    }
    for(int i = n; i >= ; --i){
        solve(str[i]);
    }
    int m = strlen(s+);
    LL ans = ;
    for(int i = ; i <= m; ++i){
        int id = s[i] - '';
        ans = (ans * base[id] + dp[id]) % mod;
    }
    cout << ans << endl;
    return ;
}