[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\n", 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[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 3e4 + ;
 const int md = ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?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 = ;

 struct BigInt {
     vi num;
     bool symbol;
     BigInt() { num.clear(); symbol = ; }
     BigInt(int x) { symbol = ; if (x < ) { symbol = ; 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 = , sum = , p = s[] == '-';
         symbol = p;
         for (int i = len - ; i >= p; i--) {
             sum += (s[i] - '') * x;
             x *= ;
             if (x == 1e8 || i == p) {
                 num.push_back(sum);
                 sum = ;
                 x = ;
             }
         }
         while (num.back() ==  && num.size() > ) 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() - ; i >= ; i--) {
             if (a[i] != b[i]) return a[i] < b[i];
         }
         return ;
     }

     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 = ;
         for (int i = ; 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() ==  && c.size() > ) c.pop_back();
         return c;
     }

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

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

     BigInt operator / (const BigInt &p) const {
         if (p == BigInt()) return p;
         BigInt res(symbol ^ p.symbol, div(num, p.num));
         if (res.symbol && res.num.size() ==  && res.num[] == ) 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() - ]);
         for (int i = num.size() - ; i >= ; i--) {
             printf("%08d", num[i]);
         }
         //putchar('\n');
     }

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

 };
 typedef BigInt bi;

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

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

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

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