题目链接：uva 1341 - Different Digits

题目大意：给定一个数字n，要求求一个数字m，m可以整除n，并且尽量组成的数字种类(0~9)尽量少，在种类相同的情况下数值尽量小。

解题思路:可以证明两种数字肯定可以组成m，假设有数字k，一种数字可以有k，kk，kkk，kkkk，…整除n剩的数一定在0~n-1之间，所以肯定存在两个由k数字组成的数字同模，那么这两个数相减所得到的数即使kkk00000，两种数字。于是肯定了范围，枚举数字，然后用bfs获取答案，维护最小值即可。

#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 65536;
int n, vis[maxn+10];

struct state {
    int key, len, vec[maxn];
    state () {
        key = 0;
    }

    bool operator < (const state& a) {
        if (a.key == 0)
            return true;
        if (key == 0)
            return false;

        if (key != a.key)
            return key < a.key;

        if (len != a.len)
            return len < a.len;

        int l = len;
        for (int i = 0; i < l; i++)
            if (vec[i] != a.vec[i])
                return vec[i] < a.vec[i];
        return false;
    }
};

struct node {
    int mod, len, num;
    void set (int mod, int num, int len) {
        this->mod = mod;
        this->num = num;
        this->len = len;
    }
}que[maxn*2];

void put (int d, state& w, int x) {
    if (d < 0 || x < 0)
        return;
    w.vec[d] = que[x].num;
    put(d-1, w, vis[que[x].mod]);
}

state judge (int a, int b) {
    int head = 1, rear = 1;
    memset(vis, -1, sizeof(vis));

    state w;
    node u, v;
    w.key = (a==b?1:2);

    if (a) {
        que[rear++].set(a%n, a, 1);
        vis[a%n] = 0;
    }

    if (b) {
        que[rear++].set(b%n, b, 1);
        vis[b%n] = 0;
    }

    while (head < rear) {
        u = que[head++];

        if (u.mod == 0) {
            w.len = u.len;
            put(u.len-1, w, head-1);
            return w;
        }

        for (int i = 0; i < w.key; i++) {
            int t = (i?b:a);
            v.set((u.mod*10+t)%n, t, u.len+1);

            if (vis[v.mod] != -1)
                continue;

            vis[v.mod] = head-1;;
            que[rear++] = v;
        }
    }
    w.key = 0;
    return w;
}

int main () {
    while (scanf("%d", &n) == 1 && n) {
        state ans;

        for (int i = 1; i <= 9; i++) {
            state c = judge(i, i);
            if (c < ans)
                ans = c;
        }

        if (ans.key == 0) {
            for (int i = 0; i <= 9; i++) {
                for (int j = i+1; j <= 9; j++) {
                    state c = judge(i, j);
                    if (c < ans)
                        ans = c;
                }
            }
        }
        for (int i = 0; i < ans.len; i++)
            printf("%d", ans.vec[i]);
        printf("\n");
    }    
    return 0;
}