Atcoder 1973：こだわり者いろはちゃん / Iroha's Obsession

C - こだわり者いろはちゃん / Iroha's Obsession

Time limit : 2sec / Memory limit : 256MB

Score : 300 points

Problem Statement

Iroha is very particular about numbers. There are K digits that she dislikes: D1,D2,…,DK.

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≦D1<D2<…<DK≦9
{D1,D2,…,DK}≠{1,2,3,4,5,6,7,8,9}

Input

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

N K
D1 D2 … DK

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

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

题意

第一行是数n和k，第二行有k个数（这k个数均在1~9之间），要求找出各位都不包含这k个数的大于n的最小的数

思路

将出现过的数字标记一下，然后for循环从n开始，写一个check函数，查找当前数字的每一位的数字是否在k个数里面，一直找到符合要求的数为止。

AC代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ull unsigned long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x7f7f7f7f
#define lson o<<1
#define rson o<<1|1
const double E=exp(1);
const int maxn=1e6+10;
const int mod=1e9+7;
using namespace std;
int a[maxn];
int vis[maxn];
bool check(int n)
{
	while(n)
	{
		if(vis[n%10])
			return false;
		n/=10;
	}
	return true;
}
int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	int n,k;
	int x;
	cin>>n>>k;
	for(int i=0;i<k;i++)
	{
		cin>>x;
		vis[x]=1;
	}
	for(int i=n;;i++)
	{
		if(check(i))
		{
			cout<<i<<endl;
			return 0;
		}
	}
	return 0;
}