Good Bye 2022 简要题解

从这里开始

• 比赛目录
• Problem A Koxia and Whiteboards
• Problem B Koxia and Permutation
• Problem C Koxia and Number Theory
• Problem D Koxia and Game
• Problem E Koxia and Game
• Problem F Koxia and Sequence
• Problem G Koxia and Bracket
• Problem H Koxia, Mahiru and Winter Festival

　　过气选手留下了只会套路的眼泪。sad......

Problem A Koxia and Whiteboards

　　相信大家都会.jpg

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#include <bits/stdc++.h> using namespace std; typedef bool boolean;   template <typename T> boolean vmin(T& a, T b) {     return (a > b) ? (a = b, true) : (false); } template <typename T> boolean vmax(T& a, T b) {     return (a < b) ? (a = b, true) : (false); }   template <typename T> T smax(T x) {     return x; } template <typename T, typename ...K> T smax(T a, const K &...args) {     return max(a, smax(args...)); }   template <typename T> T smin(T x) {     return x; } template <typename T, typename ...K> T smin(T a, const K &...args) {     return min(a, smin(args...)); }   // debugging lib   #define VN(x) #x #define Vmsg(x) VN(x) << " = " << (x) #define printv(x) cerr << VN(x) << " = " << (x); //#define debug(...) fprintf(stderr, __VA_ARGS__);   template <typename A, typename B> ostream& operator << (ostream& os, const pair<A, B>& z) {     os << "(" << z.first << ", " << z.second << ')';     return os; } template <typename T> ostream& operator << (ostream& os, const vector<T>& a) {     boolean isfirst = true;     os << "{";     for (auto z : a) {         if (!isfirst) {             os << ", ";         }         os << z;         isfirst = false;     }     os << '}';     return os; }   #define ll long long   int T; int n, m;   void solve() {   scanf("%d%d", &n, &m);   ll sum = 0;   priority_queue<int> Q;   for (int i = 1, x; i <= n; i++) {     scanf("%d", &x);     sum += x;     Q.push(-x);   }   for (int i = 1, x; i <= m; i++) {     scanf("%d", &x);     sum += Q.top() + x;     Q.pop();     Q.push(-x);   }   printf("%lld\n", sum); }   int main() {   scanf("%d", &T);   while (T--) {     solve();   }   return 0; }

Problem B Koxia and Permutation

　　当 $k = 1$ 的时候答案是 $2n$

　　当 $k > 1$ 的时候答案下界是 $n + 1$，像 $n, 1, n - 1, 2, \cdots$ 这样构造就行了

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#include <bits/stdc++.h> using namespace std;   int T, n;   void solve() {   scanf("%d%*d", &n);   int l = 1, r = n;   for (int i = 1; i <= n; i++) {     printf("%d%c", (i & 1) ? r-- : l++, i == n ? '\n' : ' ');   } }   int main() {   scanf("%d", &T);   while (T--) {     solve();   }   return 0; }

Problem C Koxia and Number Theory

　　题目相当于要使得 $1 = (a_i + x, a_j + x) = (a_i + x, a_j - a_i)$

　　假设它不是 $1$，而是 $p$ 的倍数，那么相当于要求 $x$ 模 $p$ 不能是某个值。

　　显然当 $x$ 模某个 $p$ 所有值都不能取的时候，答案为 NO，剩下都是 YES（考虑钦定模每个素数的余数 CRT 一定有解）

　　然后对 100 以内的质数暴力就可以了。

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#include <bits/stdc++.h> using namespace std;   #define ll long long   const int N = 105;   int T, n;   bitset<150> vis; vector<int> P; void get_primes() {   const int L = 140;   for (int i = 2; i < L; i++) {     if (!vis.test(i)) {       P.push_back(i);     }     for (int j = i * i; j < L; j += i) {       vis.set(j);     }   } }   ll a[N]; bool flg[150];   void solve() {   scanf("%d", &n);   for (int i = 1; i <= n; i++) {     scanf("%lld", a + i);   }   sort(a + 1, a + n + 1);   for (int i = 2; i <= n; i++) {     if (a[i] == a[i - 1]) {       puts("NO");       return;     }   }   for (auto p : P) {     for (int r = 0; r <= p; r++)       flg[r] = false;     for (int i = 1; i <= n; i++) {       for (int j = i + 1; j <= n; j++) {         ll d = a[j] - a[i];         if (d % p) {           continue;         }         flg[a[i] % p] = true;       }     }     int q = 0;     while (flg[q]) q++;     if (q >= p) {       puts("NO");       return;     }   }   puts("YES"); }   int main() {   get_primes();   scanf("%d", &T);   while (T--) {     solve();   }   return 0; }

Problem D Koxia and Game

　　容易发现 Koxia 只有一种可选项。否则考虑 Koxia 最后有选择机会的那次，剩下 Mahiru 删完都只剩 2 个一样的，如果 Koxia 能有 2 个选择，那她可以选择让它无法形成排列的那个。

　　因此每一列只有两种数，Mahiru 会删掉少的那一个，要求得到的是一个排列。

　　如果 $a_i \neq b_i$ 那么相当于要在 $a_i, b_i$ 中选择一个，如果 $a_i = b_i$ 那么一定是 $a_i$。

　　考虑建图 $a_i$ 和 $b_i$ 连一条边，表示要在这两个点中选一个。那么每个连通块需要有 $k$ 个点和 $k$ 条边，即每个连通块为基环树。

　　如果环为自环方案数为 $n$，否则为 $2$。

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#include <bits/stdc++.h> using namespace std;   #define ll long long   void exgcd(int a, int b, int& x, int& y) {     if (!b) {         x = 1, y = 0;     } else {         exgcd(b, a % b, y, x);         y -= (a / b) * x;     } }   int inv(int a, int n) {     int x, y;     exgcd(a, n, x, y);     return (x < 0) ? (x + n) : (x); }   const int Mod = 998244353;   template <const int Mod = :: Mod> class Z {     public:         int v;           Z() : v(0) {    }         Z(int x) : v(x){    }         Z(ll x) : v(x % Mod) {  }           friend Z operator + (const Z& a, const Z& b) {             int x;             return Z(((x = a.v + b.v) >= Mod) ? (x - Mod) : (x));         }         friend Z operator - (const Z& a, const Z& b) {             int x;             return Z(((x = a.v - b.v) < 0) ? (x + Mod) : (x));         }         friend Z operator * (const Z& a, const Z& b) {             return Z(a.v * 1ll * b.v);         }         friend Z operator ~(const Z& a) {             return inv(a.v, Mod);         }         friend Z operator - (const Z& a) {             return Z(0) - a;         }         Z& operator += (Z b) {             return *this = *this + b;         }         Z& operator -= (Z b) {             return *this = *this - b;         }         Z& operator *= (Z b) {             return *this = *this * b;         }         friend bool operator == (const Z& a, const Z& b) {             return a.v == b.v;         } };   Z<> qpow(Z<> a, int p) {     Z<> rt = Z<>(1), pa = a;     for ( ; p; p >>= 1, pa = pa * pa) {         if (p & 1) {             rt = rt * pa;         }     }     return rt; }   typedef Z<> Zi;   template <typename T> void pfill(T* pst, T* ped, T val) {   for ( ; pst != ped; *(pst++) = val); }   const int N = 1e5 + 5;   typedef class UnionFound {   public:     bool loop[N];     int f[N], v[N];       void init(int n) {       pfill(loop, loop + n + 1, false);       pfill(v + 1, v + n + 1, -1);       for (int i = 1; i <= n; i++) {         f[i] = i;       }     }       int find(int x) {       return f[x] == x ? x : (f[x] = find(f[x]));     }       void unit(int x, int y) {       bool flg = (x == y);       x = find(x), y = find(y);       if (x ^ y) {         f[y] = x;         v[x] += v[y];         loop[x] |= loop[y];       }       v[x]++;       loop[x] |= flg;     }       bool chk(int p) {       return v[find(p)] == 0;     }       bool has_self_loop(int p) {       return loop[find(p)];     } } UnionFound;   int T, n; int a[N], b[N]; UnionFound uf;   void solve() {   scanf("%d", &n);   for (int i = 1; i <= n; i++) {     scanf("%d", a + i);   }   for (int i = 1; i <= n; i++) {     scanf("%d", b + i);   }   uf.init(n);   for (int i = 1; i <= n; i++) {     uf.unit(a[i], b[i]);   }   for (int i = 1; i <= n; i++) {     if (!uf.chk(i)) {       puts("0");       return;     }   }   Zi ans = 1;   for (int i = 1; i <= n; i++) {     if (uf.find(i) == i) {       if (uf.loop[i]) {         ans = ans * n;       } else {         ans = ans + ans;       }     }   }   printf("%d\n", ans.v); }   int main() {   scanf("%d", &T);   while (T--) {     solve();   }   return 0; }

Problem E Koxia and Game

　　只会一个压根儿不能写的套路 dp。sad....

　　考虑每条边的贡献是两边点数的乘积。并且注意到点数改变只有两种可能。

　　考虑按顺序操作每条边的过程，每个连通块内部只会被新加入的连接它的边影响，因此只需要记录当前点 $u$ 有蝴蝶的概率就可以了。

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#include <bits/stdc++.h> using namespace std;   #define ll long long   void exgcd(int a, int b, int& x, int& y) {     if (!b) {         x = 1, y = 0;     } else {         exgcd(b, a % b, y, x);         y -= (a / b) * x;     } }   int inv(int a, int n) {     int x, y;     exgcd(a, n, x, y);     return (x < 0) ? (x + n) : (x); }   const int Mod = 998244353;   template <const int Mod = :: Mod> class Z {     public:         int v;           Z() : v(0) {    }         Z(int x) : v(x){    }         Z(ll x) : v(x % Mod) {  }           friend Z operator + (const Z& a, const Z& b) {             int x;             return Z(((x = a.v + b.v) >= Mod) ? (x - Mod) : (x));         }         friend Z operator - (const Z& a, const Z& b) {             int x;             return Z(((x = a.v - b.v) < 0) ? (x + Mod) : (x));         }         friend Z operator * (const Z& a, const Z& b) {             return Z(a.v * 1ll * b.v);         }         friend Z operator ~(const Z& a) {             return inv(a.v, Mod);         }         friend Z operator - (const Z& a) {             return Z(0) - a;         }         Z& operator += (Z b) {             return *this = *this + b;         }         Z& operator -= (Z b) {             return *this = *this - b;         }         Z& operator *= (Z b) {             return *this = *this * b;         }         friend bool operator == (const Z& a, const Z& b) {             return a.v == b.v;         } };   Z<> qpow(Z<> a, int p) {     Z<> rt = Z<>(1), pa = a;     for ( ; p; p >>= 1, pa = pa * pa) {         if (p & 1) {             rt = rt * pa;         }     }     return rt; }   typedef Z<> Zi;   template <typename T> void pfill(T* pst, T* ped, T val) {   for ( ; pst != ped; *(pst++) = val); }   const int N = 3e5 + 5;   int n, K; Zi f[N]; int eu[N], ev[N], sz[N], Fa[N]; vector<int> G[N];   int dfs(int p, int fa) {   Fa[p] = fa;   for (auto e : G[p]) {     if (e ^ fa) {       sz[p] += dfs(e, p);     }   }   return sz[p]; }   int main() {   scanf("%d%d", &n, &K);   for (int i = 1, x; i <= K; i++) {     scanf("%d", &x);     f[x] = sz[x] = 1;   }   for (int i = 1, u, v; i < n; i++) {     scanf("%d%d", &u, &v);     G[u].push_back(v);     G[v].push_back(u);     eu[i] = u, ev[i] = v;   }   dfs(1, 0);   Zi ans = 0, inv2 = (Mod + 1) >> 1;   for (int i = 1, u, v; i < n; i++) {     if (Fa[u = eu[i]] == (v = ev[i])) {       swap(u, v);     }     int l = K - sz[v], r = sz[v];     Zi puv = inv2 * f[u] * (1 - f[v]), pvu = inv2 * f[v] * (1 - f[u]);     ans += (1 - puv - pvu) * l * r;     if (l) ans += puv * (l - 1) * (r + 1);     if (r) ans += pvu * (l + 1) * (r - 1);     Zi nfu = inv2 * f[u] * (f[v] + 1) + pvu;     Zi nfv = inv2 * (f[u] + 1) * f[v] + puv;     f[u] = nfu, f[v] = nfv;   }   ans = ans * ~(Zi(K) * (K - 1) * inv2);   printf("%d\n", ans.v);   return 0; }

Problem F Koxia and Sequence

　　只会复杂度大概 1e9 的 dp，麻了。

　　考虑只要求 $a_i$ 要是 $y$ 的子集怎么算方案数。暴力的话 $\sum_{\sum a_i = x} \prod [a_i 是 y 的子集]$

　　后面那个用二进制下的组合数来表示 $\binom{y}{a_i}$，然后用生成函数或者组合恒等式容易得到这整个是 $\binom{ny}{x}$

　　算答案的话，考虑算 $a_i$ 对答案某一二进制位的贡献，由对称性，只用算 $a_1$ 就可以了。

　　然后相当于硬点 $a_1$ 包含 $2^k$，同样也能用这样的方法算方案数。

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#include <bits/stdc++.h> using namespace std;   #define ll long long   ll n, x, y, ans;   int main() {   scanf("%lld%lld%lld", &n, &x, &y);   for (int s = y; s; s = (s - 1) & y) {     // a_1 + \cdots + a_n = x (a_i \subseteq s)     // \sum \prod \binom{s}{a_i}     // ((1 + t)^s)^n [t^x]     // \binom {ns}{x}     for (int i = 0; (s >> i); i++) {       // 2^i \subseteq a_1       // \sum \binom{s-2^i}{a'_1} \prod \binom{s}{a_i}       // \binom{ns - 2^i}{x - 2^i}       ll a = n * s - (1 << i), b = x - (1 << i);       if (((s >> i) & 1) && a >= 0 && b >= 0 && (b & a) == b) {         ans ^= (1 << i);       }     }      }   ans *= n & 1;   printf("%lld\n", ans);   return 0; }

Problem G Koxia and Bracket

　　好不容易，这个我会！然后：

　　多项式题还能被卡常的吗？出题人，我*#&￥*&*&*#*&!$(*#

---

　　不难发现最优策略是：拿个栈去跑，遇到 ( push，遇到 ) 堆顶有 ( pop，否则 push，然后把栈剩下的里面的东西删掉。

　　容易证明某个被删掉的 ( 右边一定没有被删掉的 )。因此接下来我们只用考虑删 ) 的情形，剩下我们把序列翻转一下再做一遍就好了。

　　考虑栈里剩下的每个 ) 相当于是要求这个位置左侧还需要额外删一个。

　　假设总共要删 $k$ 个，我删掉了 $p_1, p_2, \cdots, p_k$ 处的 )，那么要求 $p_i$ 不能超过第 $i$ 个栈中的 ) 的位置。

　　朴素 dp 的话 $f_{i, j}$ 表示考虑到 $i$ 已经删了 $j$ 个，转移到 $f_{i + 1, j}$ 或者 $f_{i + 1, j + 1}$。

　　相当于是在网格图里向右或者向右上走一格，这个右上非常地不优美，把第 $k$ 行向左推 $k - 1$ 格就变成向右或者向上走一格。

　　然后就变成在一个梯形的网格图上路径计数。

　　注意到如果是一个优美的矩形，算方案数显然能直接卷积。

　　所以考虑分治

　　对于矩形部分用卷积算，剩下两块阴影部分分别拿去递归。

　　因为矩形算卷积是已知刚跨过左边界和下边界上每条边的方案数，然后算刚跨过右边界和上边界上每条边的方案数，要做 4 次卷积，就被卡了。当矩形长宽不大的时候，直接把这部分换成平方 dp 就能过了。

　　题解做法的话是记录额外删了多少个，限制的话变成要求额外删的数量总是大于等于 0 的。

　　假如现在需要快速转移 $[l, r]$，假设这里面有 $num$ 个需要被删的 )，那么当第二维超过 $num$ 的时候转移不会有任何特殊影响，直接用卷积算出对 $r + 1$ 的贡献就可以了。

　　所以一个区间实际需要在内部维护的 dp 第二维只和这里面需要被删的 ) 有关。因此直接分治就好了。

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#include <bits/stdc++.h> using namespace std;   #define g __ntt_g   #ifdef local #define _assert(expr) assert(expr) #else #define _assert(expr) #endif   const int Mod = 998244353; const int bzmax = 20; const int N = 1 << bzmax; const int g = 3;   typedef long long ll;   void exgcd(int a, int b, int& x, int& y) {   if (!b) {     x = 1, y = 0;   } else {     exgcd(b, a % b, y, x);     y -= (a / b) * x;   } } int inv(int a, int Mod = ::Mod) {   int x, y;   exgcd(a, Mod, x, y);   return (x < 0) ? (x + Mod) : x; }   template <const int Mod = ::Mod> class Z {   public:     int v;       Z() {   }     Z(int v) : v(v) {       _assert(v >= 0 && v < Mod);     }     Z(ll x) : v(x % Mod) {  }       friend Z operator + (Z a, Z b) {       int x;       return Z((x = a.v + b.v) >= Mod ? x - Mod : x);     }     friend Z operator - (Z a, Z b) {       int x;       return Z((x = a.v - b.v) < 0 ? x + Mod : x);     }     friend Z operator * (Z a, Z b) {       return 1ll * a.v * b.v;     }     friend Z operator ~ (Z a) {       _assert(a.v);       return inv(a.v);     }     friend Z operator - (Z a) {       return Z(0) - a;     }     Z& operator += (Z b) {       return *this = *this + b;     }     Z& operator -= (Z b) {       return *this = *this - b;     }     Z& operator *= (Z b) {       return *this = *this * b;     } };   typedef Z<> Zi;   Zi qpow(Zi a, int p) {   if (p < 0)     p += Mod - 1;   Zi rt = 1;   for ( ; p; p >>= 1, a *= a) {     if (p & 1) {       rt *= a;     }   }   return rt; } Zi qpow(Zi a, ll p) {   return qpow(a, (int) (p % (Mod - 1))); }   class NTT {   private:     Zi gn[bzmax + 1], _gn[bzmax + 1];     Zi Wn[2][N << 1];     public:     NTT() {       int phi = Mod - 1;       for (int i = 0; i <= bzmax; i++) {         gn[i] = qpow(g, (phi >> i));         _gn[i] = ~gn[i];       }       Zi *cw = Wn[0];       for (int i = 1; i <= bzmax; i++) {         *(cw++) = 1;         for (int j = (1 << (i - 1)) - 1; j--; cw++) {           *cw = *(cw - 1) * gn[i];         }       }       cw = Wn[1];       for (int i = 1; i <= bzmax; i++) {         *(cw++) = 1;         for (int j = (1 << (i - 1)) - 1; j--; cw++) {           *cw = *(cw - 1) * _gn[i];         }       }     }       void operator () (Zi* f, int len, bool rev) {       _assert(len <= N);       for (int i = 1, j = len >> 1, k; i < len - 1; i++, j += k) {         if (i < j) swap(f[i], f[j]);         for (k = len >> 1; j >= k; j -= k, k >>= 1);       }       Zi *wn = Wn[rev], *cw, a, b, *x, *y;       for (int l = 2, hl; l <= len; l <<= 1) {         hl = l >> 1, cw = wn;         for (int i = 0; i < len; i += l) {           wn = cw, x = f + i, y = x + hl;           for (int j = hl; j--; ) {             a = *x, b = *y * *(wn++);             *(x++) = a + b, *(y++) = a - b;           }         }       }       if (rev) {         a = ~Zi(len);         for ( ; len--; *(f++) *= a);       }     } } NTT;   #define dft(f, len) NTT(f, len, false) #define idft(f, len) NTT(f, len, true) #define clr(f, len) memset(f, 0, (len) << 2)   int get_dft_length(int len) {   int rt = 1;   while (rt < len)     rt <<= 1;   return rt; }   #define get_length(_x) get_dft_length(_x)   #undef g   template <typename T1, typename T2> void do_add(T1* a, const T2 * b, int n) {   while (n--) *(a++) += *(b++); } template <typename T1, typename T2> void do_sub(T1* a, const T2 * b, int n) {   while (n--) *(a++) -= *(b++); } template <typename T1, typename T2> void do_mul(T1* a, const T2 * b, int n) {   while (n--) *(a++) *= *(b++); }   namespace poly_temporary {   Zi ta[N], tb[N]; }   void poly_inverse(const Zi *f, Zi* g, int n) {   using namespace poly_temporary;   static Zi A[N];   if (n > 64) {     //  if (n > 1) {     int hn = (n + 1) >> 1;     int t = get_length(hn * 3);     poly_inverse(f, g, hn);       copy(f, f + n, A);     clr(A + n, t - n);     copy(g, g + hn, ta);     dft(g, t);     dft(A, t);     for (int i = 0; i < t; i++) {       g[i] = g[i] * (Zi(2) - g[i] * A[i]);     }     idft(g, t);     copy(ta, ta + hn, g);     clr(g + n, t - n);   } else {     g[0] = ~f[0];     for (int i = 1; i < n; i++) {       g[i] = 0;       for (int j = 1; j <= i; j++) {         g[i] -= f[j] * g[i - j];       }       g[i] *= g[0];     }   } }   vector<Zi> fac, _fac, Inv;   void init_fac(int n) {   fac.resize(n + 1);   _fac.resize(n + 1);   fac[0] = 1;   for (int i = 1; i <= n; i++) {     fac[i] = fac[i - 1] * i;   }   _fac[n] = ~fac[n];   for (int i = n; i; i--) {     _fac[i - 1] = _fac[i] * i;   } } void init_inv(int n) {   Inv.resize(n + 1);   Inv[0] = 0, Inv[1] = 1;   for (int i = 2; i <= n; i++) {     Inv[i] = -Inv[Mod % i] * (Mod / i);   } }   typedef class Poly : public vector<Zi> {   public:     using vector<Zi>::vector;       Poly& fix(int sz) {       this->resize(sz, 0);       return *this;     }       Zi eval(Zi x) const {       Zi rt = 0;       const Zi *f = data();       for (int i = size(); i; rt = rt * x + f[--i]);       return rt;     }     vector<Zi> eval(Zi* x, int n); // TODO     vector<Zi> eval(vector<Zi> x) {       return eval(x.data(), x.size());     }       void shrink() {       while (size() > 1u && !back().v)         pop_back();     }       Poly deri();     Poly integ();     Poly& do_deri();     Poly& do_integ(); } Poly;   ostream& operator << (ostream& os, const Poly& f) {   bool first = true;   os << "{";   for (auto x : f) {     if (first) {       first = false;     } else {       os << ", ";     }     os << x.v;   }   os << "}";   return os; }   Poly operator + (const Poly& a, const Poly& b) {   Poly rt = a;   int n = max(a.size(), b.size());   rt.resize(n);   for (int i = b.size(); i--; )     rt[i] += b[i];   return rt; } Poly operator - (const Poly& a, const Poly& b) {   Poly rt = a;   int n = max(a.size(), b.size());   rt.resize(n);   for (int i = b.size(); i--; )     rt[i] -= b[i];   return rt; }   Poly operator * (Poly a, Poly b) {   int n = a.size(), m = b.size(), k = n + m - 1;   if (n < 64 || m < 64) {     Poly rt (k, Zi(0));     for (int i = a.size(); i--; ) {       for (int j = b.size(); j--; ) {         rt[i + j] += a[i] * b[j];       }     }     return rt;   }   int t = get_length(k);   a.resize(t, Zi(0));   b.resize(t, Zi(0));   dft(a.data(), t);   dft(b.data(), t);   do_mul(a.data(), b.data(), t);   idft(a.data(), t);   return a.fix(k); }   Poly operator ~ (Poly b) {   int n = b.size(), t = get_length((n << 1) + (n & 1));   Poly rt (t, Zi(0));   poly_inverse(b.data(), rt.data(), n);   return rt.fix(n); }   Poly operator / (Poly a, Poly b) {   int n = a.size(), m = b.size();   if (n < m) return Poly {0};   int d = n - m + 1;   reverse(a.begin(), a.end());   reverse(b.begin(), b.end());   a.resize(d, Zi(0));   b.resize(d, Zi(0));   a = (a * ~b).fix(d);   reverse(a.begin(), a.end());   return a; }   Poly operator % (Poly a, Poly b) {   int n = a.size(), m = b.size();   if (n < m) return a;   if (!m) return Poly{0};   return (a - a / b * b).fix(m - 1); }   pair<Poly, Poly> divide(Poly a, Poly b) {   int n = a.size(), m = b.size();   if (n < m) return make_pair(Poly{0}, a);   if (!m) return make_pair(a / b, Poly{0});   Poly d = a / b;   return make_pair(d, (a - d * b).fix(m - 1)); }   Poly Poly::deri() {   Poly b = *this;   return b.do_deri(); } Poly& Poly::do_deri() {   shrink();   Zi* f = data();   int sz = size();   for (int i = 0; i < sz - 1; i++) {     f[i] = f[i + 1] * (i + 1);   }   f[sz - 1] = 0;   if (sz > 1) {     pop_back();   }   return *this; }   Poly Poly::integ() {   Poly b = *this;   return b.do_integ(); } Poly& Poly::do_integ() {   push_back(0);   Zi* f = data();   int sz = size();   _assert(sz < (signed) Inv.size());   for (int i = sz; --i; ) {     f[i] = f[i - 1] * Inv[i];   }   f[0] = 0;   return *this; }   Poly ln(Poly a) {   int n = a.size();   _assert(!a.empty() && a[0].v == 1);   return (~a * a.deri()).fix(n - 1).integ(); }   void _exp(Poly& f, Poly& g, int n) { //  using namespace poly_temporary;   if (n == 1) {     g = Poly{1};   } else {     int hn = (n + 1) >> 1;     _exp(f, g, hn);           Poly gn = g;     Poly h = Poly(f.begin(), f.begin() + n) - ln(gn.fix(n));     h.erase(h.begin(), h.begin() + hn);     h = g * h;     g.resize(n);     for (int i = hn; i < n; i++) {       g[i] = h[i - hn];     }   } } Poly exp(Poly a) {   _assert(a[0].v == 0);   Poly h;   _exp(a, h, a.size());   return h; }   #undef clr #undef get_length   int L; char s[N];   void do_reverse() {   char *p = s;   for (; *p; p++) {     *p = "()"[*p == '('];   }   reverse(s, p); }   Poly subpoly(Poly& f, int l, int r) {   Poly g (r - l);   for (int i = l; i < r; i++) {     g[i - l] = f[i];   }   return g; }   Poly revpoly(Poly f) {   reverse(f.begin(), f.end());   return f; }   Poly eval_cross(Poly f, int m) {   int n = f.size();   for (int i = 0; i < n; i++) {     f[i] *= _fac[n - 1 - i];   }   Poly g (n + m - 1);   for (int i = 0; i < (signed) g.size(); i++) {     g[i] = fac[i];   }   f = f * g;   g.resize(m);   for (int i = 0; i < m; i++) {     g[i] = f[i + n - 1] * _fac[i];   }   return g; }   Poly eval_paral(Poly f, int m) {   int n = f.size();   Poly g (n);   for (int i = 0; i < n; i++) {     g[i] = fac[i + m] * _fac[i] * _fac[m];   }   (f = f * g).resize(n);   return f; }   Poly join(Poly f, Poly g) {   int n = f.size(), m = g.size();   f.resize(n + m);   for (int i = 0; i < m; i++) {     f[i + n] = g[i];   }   return f; }   vector<pair<int, int>> ps; Poly solve(int pl, int pr, Poly f) {   if (pl == pr) {     return eval_cross(f, ps[pr + 1].first - ps[pl].first);   }   int mid = (pl + pr) >> 1;   int wid = ps[mid + 1].first - ps[pl].first;   int hei = ps[pr + 1].second - ps[mid + 1].second;     Poly h = solve(pl, mid, subpoly(f, 0, ps[mid + 1].second - ps[pl].second));   Poly nf, R0;   f = subpoly(f, ps[mid + 1].second - ps[pl].second, f.size());   if (wid < 256 || hei < 256) {     nf.resize(hei, 0);     R0.resize(wid, 0);     for (int i = 0; i < hei; i++) {       for (int j = 1; j < wid; j++) {         h[j] += h[j - 1];       }       nf[i] += h.back();     }     R0 = h;     for (int i = 0; i < wid; i++) {       for (int j = 1; j < hei; j++) {         f[j] += f[j - 1];       }       R0[i] += f.back();     }     nf = nf + f;   } else {     nf = eval_cross(h, hei) + eval_paral(f, wid - 1);     R0 = eval_cross(f, wid) + eval_paral(h, hei - 1);   }   Poly R1 = solve(mid + 1, pr, nf);   return join(R0, R1); }   Zi solve() {   vector<int> pos;   int tp = 0;   int cnt = 0;   for (char *p = s; *p; p++) {     if (*p == '(') {       tp++;     } else {       cnt++;       !tp ? (pos.push_back(cnt), 0) : tp--;     }   }   if (pos.empty()) {     return 1;   }   for (int i = 1; i < (signed) pos.size(); pos[i] -= i, i++);   ps.clear();   ps.emplace_back(0, 1);   for (int i = 1; i < (signed) pos.size(); i++) {     if (pos[i] != pos[i - 1]) {       ps.emplace_back(pos[i - 1], i + 1);     }   }   ps.emplace_back(pos.back(), (int) pos.size() + 1);   Poly fi (pos.size(), 0);   fi[0] = 1;   Poly f = solve(0, (signed) ps.size() - 2, fi);   Zi ret = 0;   for (auto x : f) {     ret += x;   }   return ret; }   int main() {   scanf("%s", s);   int len_s = strlen(s);   init_fac(len_s + 1);   Zi a = solve();   do_reverse();   Zi b = solve();   printf("%d", (a * b).v);   return 0; }

Problem H Koxia, Mahiru and Winter Festival

别催了，别催了，在路上了.jpg