Problem:
Score : 300 points

Problem Statement

Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K.

She is shopping, and now paying at the cashier.
Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change).

However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money.

Find the amount of money that she will hand to the cashier.

Constraints

1 ≦ N < 10000

1 ≦ K < 10

0 ≦ D_1 < D_2 < … < D_K≦9

\{D_1,D_2,...,D_K\} ≠ \{1,2,3,4,5,6,7,8,9\}

Input

The input is given from Standard Input in the following format:

N K
D_1 D_2 … D_K

Output

Print the amount of money that Iroha will hand to the cashier.

Sample Input 1

1000 8
1 3 4 5 6 7 8 9

Sample Output 1

2000

She dislikes all digits except 0 and 2.

The smallest integer equal to or greater than N=1000 whose decimal notation contains only 0 and 2, is 2000.

Sample Input 2

9999 1
0

Sample Output 2

9999

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;

/*
 <url:https://arc058.contest.atcoder.jp/tasks/arc058_a>
 問題文============================================================
 いろはちゃんはこだわりもので、嫌いな数字が K 個あり、それぞれ D1,D2,…,DK です。
 
 いろはちゃんはお店でお買い物をしていて、 N 円の品物を買おうとしています。
 もちろん、この品物は N 円以上のお金を支払えば買うことができます。
 しかし、先ほど述べたようにいろはちゃんは強いこだわりがあるので、
 自分がお店に支払う金額の 10 進表記にいろはちゃんの嫌いな数字が出現しないような最も少ない金額を支払おうとします。
 
 いろはちゃんが支払う金額を求めてください。
 =================================================================
 
 解説=============================================================
 
 ================================================================
 */
int main(void) {
    cin.tie(0); ios::sync_with_stdio(false);
    string N; int K; cin >> N >> K;
    bool D[10];
    for(int i = 0; i < K;i++){
        int x; cin >> x;
        D[x] = true;
    }
    int minv = 100;
    for(int i = 1; i < 10; i++) if(!D[i]) minv = min(minv,i);
    
    string ans = "";
    for(int i = 0; i < (int)N.length();i++){
        bool f = false;
        for(int j = 0; j < 10;j++){
            if(D[j]) continue;
            if(j >= (N[i] - '0')){
                ans += to_string(j);
                f = true;
                break;
            }
        }
        if(!f){
            ans += to_string(minv);
            for(int j = i; j < (int)N.length();j++){
                if(D[0]) ans += to_string(minv);
                else ans += "0";
            }
            break;
        }
    }
    cout << ans << endl;
    return 0;
}