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;
}