Yet Another Multiple Problem（同餘方程+BFS）

SP12810

There are tons of problems about integer multiples. Despite the fact that the topic is not original, the content is highly challenging. That’s why we call it “Yet Another Multiple Problem”.
In this problem, you’re asked to solve the following question: Given a positive integer n and m decimal digits, what is the minimal positive multiple of n whose decimal notation does not contain any of the given digits?
Input
There are several test cases.
For each test case, there are two lines. The first line contains two integers n and m (1 ≤ n ≤ 10
4). The second line contains m decimal digits separated by spaces.
Input is terminated by EOF.
Output
For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) while Y is the minimal multiple satisfying the above-mentioned conditions or “-1” (without quotation marks) in case there does not exist such a multiple.
Sample Input
2345 3
7 8 9
100 1
0
Sample Output
Case 1: 2345
Case 2: -1

將合法的數位保存下來，然後將合法的數爲拼接起來，判斷是否是$n$的倍數。
採用$BFS$能保證搜索的數字是遞增的，保證第一個找到的是最優解。
設兩個數$A,B$，若$A\equiv B\pmod n$那麼將數位$C$拼接到$A,B$後面，有$AC\equiv BC\pmod n$，簡單的證明如下：

因爲：
$AC\mod n\\=(10*A+C)\mod n\\=(10*A\mod b+C\mod n)\mod n\\=((10\mod n)(A\mod n)\mod n+C\mod n)\mod n$同理：
$BC\mod n\\=((10\mod n)(B\mod n)\mod n+C\mod n)\mod n$
由$A\equiv B\pmod n$，得：$AC\equiv BC\pmod n$。
證畢。

因此如果有$A<B$且$A\equiv B\pmod n$，那麼$B$就不用搜了，因爲$B$在拼接後的所有結果在模$n$的結果都被$A$搜索過了，因此可以做一個剪枝。

由於模$n$的每一個結果都只搜索一次，事件複雜度爲$O(n)$。

數爲拼接後模數的更新：
$AC\mod n=(10*A+C)\mod n$

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <cstdio>
#include <cstring>
#include <queue>
#include <string>
using namespace std;

int n, m;
vector<int> Digits;
bool Vis[10005];

struct Node {
	int mod;
	string Number;
};

inline bool Input() {

	memset(Vis, false, sizeof(Vis));
	Digits.clear();

	bool No[10];
	memset(No, false, sizeof(No));
	if (scanf("%d%d", &n, &m) == EOF) {
		return false;
	}
	while (m--) {
		int x;
		scanf("%d", &x);
		No[x] = true;
	}
	for (int i = 0; i < 10; ++i) {
		if (No[i] == false) {
			Digits.push_back(i);
		}
	}
	return true;
}

void BFS() {
	queue<Node>Q;
	for (int i = 0; i < Digits.size(); ++i) {
		const int& digit = Digits[i];
		if (digit == 0) {
			continue;
		}
		Node node;
		node.mod = digit % n;
		node.Number = digit + '0';

		if (!node.mod) {
			cout << node.Number << endl;
			return;
		}
		Q.push(node);
	}

	while (!Q.empty()) {
		Node CurNode = Q.front();
		Q.pop();
		for (int i = 0; i < Digits.size(); ++i) {
			const int& digit = Digits[i];
			Node NewNode;
			string a;
			a = digit + '0';

			NewNode.Number = CurNode.Number + a;
			NewNode.mod = (CurNode.mod * 10 + digit) % n;
			if (!NewNode.mod) {
				cout << NewNode.Number << endl;
				return;
			}
			else if (Vis[NewNode.mod] == false) {
				Q.push(NewNode);
				Vis[NewNode.mod] = true;
			}
		}
	}
	cout << "-1" << endl;
	return;
}

int main() {
	int Case = 0;
	while (Input()) {

		cout << string("Case ") << ++Case << string(": ");
		BFS();
	}

	return 0;

}