ZOJ - 1136 Multiple (同余+BFS)

Description

a program that, given a natural number N between 0 and 4999 (inclusively), and M distinct decimal digits X1,X2..XM (at least one), finds the smallest strictly positive multiple of N that has no other digits besides X1,X2..XM (if such a multiple exists).

The input file has several data sets separated by an empty line, each data set having the following format:

On the first line - the number N
On the second line - the number M
On the following M lines - the digits X1,X2..XM.

For each data set, the program should write to standard output on a single line the multiple, if such a multiple exists, and 0 otherwise.

An example of input and output:

Input

22
3
7
0
1

2
1
1

Output

110

0

题意：让你用m个数组成n的倍数，要求最小

思路：首先我们排序这m个数，然后从高位开始搜，这样才能保证最先BFS的是最小的，这里用到了一个剪枝就是对于：余数相同的，我们可以拿来判重，小的比较优，也就是最先放进队列的比较优

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 10005;

struct Node {
	int res;
	int pre;
	int digit;
} q[maxn];
int a[maxn], vis[maxn];
int n, m;

void print(int u) {
	if (q[u].pre == -1)
		return;
	print(q[u].pre);
	printf("%d", q[u].digit);
}

int bfs() {
	q[0].digit = 0;
	q[0].pre = -1;
	q[0].res = 0;
	memset(vis, 0, sizeof(vis));
	int front = 0, rear = 1;
	while (front < rear) {
		Node tmp = q[front];
		for (int i = 0; i < m; i++) {
			int t = (tmp.res * 10 + a[i]) % n;
			if (a[i] == 0 && tmp.pre == -1)
				continue;
			if (!vis[t]) {
				vis[t] = 1;
				Node tmp;
				tmp.digit = a[i];
				tmp.pre = front;
				tmp.res = t;
				q[rear++] = tmp;
			}
			if (t == 0) {
				print(rear-1);
				printf("\n");
				return 1;
			}
		}
		front++;
	}
	return 0;
}

int main() {
	while (scanf("%d%d", &n, &m) != EOF) {
		for (int i = 0; i < m; i++)
			scanf("%d", &a[i]);
		if (n == 0) {
			printf("0\n");
			continue;
		}
		sort(a, a+m);
		if (!bfs())
			printf("0\n");
	}
	return 0;
}