[zoj3596]DP(BFS)

题意：求n的最小倍数，满足性质P：十进制的每一位上的数有m种（0<m<=10）。

思路：直接枚举n的最小倍数，然后检测是否满足性质P，n一大很容易超时，并且无法判断无解的情况。巧妙的做法是一位位构造数，同时保存每种数字的使用情况和对n的余数作为状态，如果一个状态出现过，那么后面再一次出现的时候位数肯定比之前多，故答案没之前优，舍弃。由于取模的性质，转移方程也很好写，具体见代码。

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt5(name, a, b, c, d, e) name(int a = 0, int b = 0, int c = 0, int d = 0, int e = 0): a(a), b(b), c(c), d(d), e(e) {}
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a)
#define sint(a) scanf("%d", &a)
#define sint2(a, b) scanf("%d%d", &a, &b)
#define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pint(a) printf("%d
", a)
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
#define mp(a, b) make_pair(a, b)
#define pb(a) push_back(a)

typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 3e4 + 7;
const int md = 10007;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

const int maxI = 1e8;
const int Len = 8;

struct BigInt {
    vi num;
    bool symbol;
    BigInt() { num.clear(); symbol = 0; }
    BigInt(int x) { symbol = 0; if (x < 0) { symbol = 1; x = -x; } num.push_back(x % maxI); if (x >= maxI) num.push_back(x / maxI); }
    BigInt(bool s, vi x) { symbol = s;  num = x; }
    BigInt(char s[]) {
        int len = strlen(s), x = 1, sum = 0, p = s[0] == '-';
        symbol = p;
        for (int i = len - 1; i >= p; i--) {
            sum += (s[i] - '0') * x;
            x *= 10;
            if (x == 1e8 || i == p) {
                num.push_back(sum);
                sum = 0;
                x = 1;
            }
        }
        while (num.back() == 0 && num.size() > 1) num.pop_back();
    }

    void push(int x) { num.push_back(x); }

    BigInt abs() const { return BigInt(false, num); }

    bool smaller(const vi &a, const vi &b) const {
        if (a.size() != b.size()) return a.size() < b.size();
        for (int i = a.size() - 1; i >= 0; i--) {
            if (a[i] != b[i]) return a[i] < b[i];
        }
        return 0;
    }

    bool operator < (const BigInt &p) const {
        if (symbol && !p.symbol) return true;
        if (!symbol && p.symbol) return false;
        if (symbol && p.symbol) return smaller(p.num, num);
        return smaller(num, p.num);
    }

    bool operator > (const BigInt &p) const {
        return p < *this;
    }

    bool operator == (const BigInt &p) const {
        return !(p < *this) && !(*this < p);
    }

    bool operator >= (const BigInt &p) const {
        return !(*this < p);
    }

    bool operator <= (const BigInt &p) const {
        return !(p < *this);
    }

    vi add(const vi &a, const vi &b) const {
        vi c;
        c.clear();
        int x = 0;
        for (int i = 0; i < a.size(); i++) {
            x += a[i];
            if (i < b.size()) x += b[i];
            c.push_back(x % maxI);
            x /= maxI;
        }
        for (int i = a.size(); i < b.size(); i++) {
            x += b[i];
            c.push_back(x % maxI);
            x /= maxI;
        }
        if (x) c.push_back(x);
        while (c.back() == 0 && c.size() > 1) c.pop_back();
        return c;
    }

    vi sub(const vi &a, const vi &b) const {
        vi c;
        c.clear();
        int x = 1;
        for (int i = 0; i < b.size(); i++) {
            x += maxI + a[i] - b[i] - 1;
            c.push_back(x % maxI);
            x /= maxI;
        }
        for (int i = b.size(); i < a.size(); i++) {
            x += maxI + a[i] - 1;
            c.push_back(x % maxI);
            x /= maxI;
        }
        while (c.back() == 0 && c.size() > 1) c.pop_back();
        return c;
    }

    vi mul(const vi &a, const vi &b) const {
        vi c;
        c.resize(a.size() + b.size());
        for (int i = 0; i < a.size(); i++) {
            for (int j = 0; j < b.size(); j++) {
                LL tmp = (LL)a[i] * b[j] + c[i + j];
                c[i + j + 1] += tmp / maxI;
                c[i + j] = tmp % maxI;
            }
        }
        while (c.back() == 0 && c.size() > 1) c.pop_back();
        return c;
    }

    vi div(const vi &a, const vi &b) const {
        vi c(a.size()), x(1, 0), y(1, 0), z(1, 0), t(1, 0);
        y.push_back(1);
        for (int i = a.size() - 1; i >= 0; i--) {
            z[0] = a[i];
            x = add(mul(x, y), z);
            if (smaller(x, b)) continue;
            int l = 1, r = maxI - 1;
            while (l < r) {
                int m = (l + r + 1) >> 1;
                t[0] = m;
                if (smaller(x, mul(b, t))) r = m - 1;
                else l = m;
            }
            c[i] = l;
            t[0] = l;
            x = sub(x, mul(b, t));
        }
        while (c.back() == 0 && c.size() > 1) c.pop_back();
        return c;
    }

    BigInt operator + (const BigInt &p) const{
        if (!symbol && !p.symbol) return BigInt(false, add(num, p.num));
        if (!symbol && p.symbol) return *this >= p.abs()? BigInt(false, sub(num, p.num)) : BigInt(true, sub(p.num, num));
        if (symbol && !p.symbol) return (*this).abs() > p? BigInt(true, sub(num, p.num)) : BigInt(false, sub(p.num, num));
        return BigInt(true, add(num, p.num));
    }

    BigInt operator - (const BigInt &p) const {
        return *this + BigInt(!p.symbol, p.num);
    }

    BigInt operator * (const BigInt &p) const {
        BigInt res(symbol ^ p.symbol, mul(num, p.num));
        if (res.symbol && res.num.size() == 1 && res.num[0] == 0) res.symbol = false;
        return res;
    }

    BigInt operator / (const BigInt &p) const {
        if (p == BigInt(0)) return p;
        BigInt res(symbol ^ p.symbol, div(num, p.num));
        if (res.symbol && res.num.size() == 1 && res.num[0] == 0) res.symbol = false;
        return res;
    }

    BigInt operator % (const BigInt &p) const {
        return *this - *this / p * p;
    }

    void show() const {
        if (symbol) putchar('-');
        printf("%d", num[num.size() - 1]);
        for (int i = num.size() - 2; i >= 0; i--) {
            printf("%08d", num[i]);
        }
        //putchar('
');
    }

    int TotalDigit() const {
        int x = num[num.size() - 1] / 10, t = 1;
        while (x) {
            x /= 10;
            t++;
        }
        return t + (num.size() - 1) * Len;
    }

};
typedef BigInt bi;

struct Node {
    int used, rest, cnt, fa, val;
    constructInt5(Node, used, rest, cnt, fa, val);
} que[3000000];
bool mark[1 << 10][1000];
int n, m;

bi get(int pos) {
    if (pos == 0) return 0;
    return get(que[pos].fa) * 10 + que[pos].val;
}

void bfs() {
    mem0(mark);
    int head = 0, tail = 0;
    que[tail ++] = Node(0, 0, 0, 0, 0);
    Node hnode;
    while (head < tail) {
        hnode = que[head ++];
        if (hnode.cnt > m) continue;
        if (hnode.cnt == m && hnode.rest == 0) break;
        rep_up0(i, 10) {
            if (hnode.cnt == 0 && i == 0) continue;
            if (hnode.used & (1 << i)) {
                Node newn = Node(hnode.used, (hnode.rest * 10 + i) % n, hnode.cnt, head - 1, i);
                if (mark[newn.used][newn.rest]) continue;
                mark[newn.used][newn.rest] = true;
                que[tail ++] = newn;
            }
            else {
                Node newn = Node(hnode.used | (1 << i), (hnode.rest * 10 + i) % n, hnode.cnt + 1, head - 1, i);
                if (mark[newn.used][newn.rest]) continue;
                mark[newn.used][newn.rest] = true;
                que[tail ++] = newn;
            }
        }
    }
    if (hnode.cnt != m || hnode.rest != 0) {
        puts("Impossible");
        return ;
    }
    bi ans = get(head - 1);
    ans.show();
    printf("=%d*", n);
    (ans / n).show();
    pchr('
');
}

int main() {
    //freopen("in.txt", "r", stdin);
    int T;
    cin >> T;
    while (T --) {
        cin >> n >> m;
        bfs();
    }
    return 0;
}