2023牛客第七场补题报告C F L M

2023牛客第七场补题报告C F L M

C-Beautiful Sequence_2023牛客暑期多校训练营7 (nowcoder.com)

思路

观察到数组一定是递增的，所以从最高位往下考虑每位的1最多只有一个，然后按位枚举贪心即可。

代码

#include <bits/stdc++.h>
using namespace std;
#define int long long
void solve();
signed main(){cin.sync_with_stdio(0);cin.tie(0);int T = 1;cin >> T;while(T--){solve();}return 0;
}const int N = 1e6 + 5;
bool bit[N][32];
int nok[32][2];
void dfs(int l, int r, int d){//cout << " " << d << " " << l << " " << r << "\n";if(d < 0 || l >= r)return;int conv = 0;int id = l - 1;for(int i = l;i < r;i++){if(bit[i][d] != bit[i + 1][d]){id = i;conv++;}}if(conv > 1){nok[d][0] = 1;nok[d][1] = 1;return;}	else if(conv == 1){if(bit[l][d] == 0){nok[d][1] = 1;}else{nok[d][0] = 1;}dfs(l, id, d - 1);dfs(id + 1, r, d - 1);}else{dfs(l, r, d - 1);}}
void solve(){int n,k;cin >> n >> k;vector<int> b(n + 1);for(int i = 1;i < n;i ++) {cin >> b[i];} for(int i = 0;i <= 30;i++){nok[i][0] = 0;nok[i][1] = 0;}for(int i = 1;i <= n;i ++) {for(int j = 0;j <= 30;j ++) {if(i == 1){bit[i][j] = 0;}else{bit[i][j] = (bit[i - 1][j] ^ ((b[i - 1] >> j) & 1));}}}dfs(1, n, 30);for(int i = 0;i < 30;i++){if(nok[i][0] == 1 && nok[i][1] == 1){cout << "-1\n";return;}}long long rag = 1;for(int i = 0;i <= 30;i++){if(nok[i][0] == 0 && nok[i][1] == 0){rag <<= 1;}}if(rag < k){cout << "-1\n";}else{k--;long long ans = 0;for(int i = 0, kk = 0;i <= 30;i++){if(nok[i][0] == 0 && nok[i][1] == 0){if((k >> kk) & 1){ans |= 1 << i;}kk++;}else if(nok[i][1] == 0){ans |= 1 << i;}}if(ans >= (1ll << 30)){cout << "-1\n";}else{cout << ans << " ";for(int i = 1;i < n;i++){ans = ans ^ b[i];cout << ans << " ";}cout << "\n";}}
}

F-Counting Sequences_2023牛客暑期多校训练营7 (nowcoder.com)

思路

考虑到数据范围很大，先想的dp会t，可以使用多项式乘法，n次幂乘以m次幂就可以变换为n+m，此时直接输出多项式幂次的第k项即可。

代码

#include <bits/stdc++.h>
using namespace std;void solve();int main(){cin.sync_with_stdio(0);cin.tie(0);int T = 1;//cin >> T;while(T--){solve();}return 0;
}constexpr int P = 998244353;
using i64 = long long;
// assume -P <= x < 2P
int norm(int x) {if (x < 0) {x += P;}if (x >= P) {x -= P;}return x;
}
template<class T>
T power(T a, int b) {T res = 1;for (; b; b /= 2, a *= a) {if (b % 2) {res *= a;}}return res;
}struct Z {int x;Z(int x = 0) : x(norm(x)) {}int val() const {return x;}Z operator-() const {return Z(norm(P - x));}Z inv() const {assert(x != 0);return power(*this, P - 2);}Z &operator*=(const Z &rhs) {x = i64(x) * rhs.x % P;return *this;}Z &operator+=(const Z &rhs) {x = norm(x + rhs.x);return *this;}Z &operator-=(const Z &rhs) {x = norm(x - rhs.x);return *this;}Z &operator/=(const Z &rhs) {return *this *= rhs.inv();}friend Z operator*(const Z &lhs, const Z &rhs) {Z res = lhs;res *= rhs;return res;}friend Z operator+(const Z &lhs, const Z &rhs) {Z res = lhs;res += rhs;return res;}friend Z operator-(const Z &lhs, const Z &rhs) {Z res = lhs;res -= rhs;return res;}friend Z operator/(const Z &lhs, const Z &rhs) {Z res = lhs;res /= rhs;return res;}friend std::istream &operator>>(std::istream &is, Z &a) {i64 v;is >> v;a = Z(v);return is;}friend std::ostream &operator<<(std::ostream &os, const Z &a) {return os << a.val();}
};std::vector<int> rev;
std::vector<Z> roots{0, 1};
void dft(std::vector<Z> &a) {int n = a.size();if (int(rev.size()) != n) {int k = __builtin_ctz(n) - 1;rev.resize(n);for (int i = 0; i < n; i++) {rev[i] = rev[i >> 1] >> 1 | (i & 1) << k;}}for (int i = 0; i < n; i++) {if (rev[i] < i) {std::swap(a[i], a[rev[i]]);}}if (int(roots.size()) < n) {int k = __builtin_ctz(roots.size());roots.resize(n);while ((1 << k) < n) {Z e = power(Z(3), (P - 1) >> (k + 1));for (int i = 1 << (k - 1); i < (1 << k); i++) {roots[2 * i] = roots[i];roots[2 * i + 1] = roots[i] * e;}k++;}}for (int k = 1; k < n; k *= 2) {for (int i = 0; i < n; i += 2 * k) {for (int j = 0; j < k; j++) {Z u = a[i + j];Z v = a[i + j + k] * roots[k + j];a[i + j] = u + v;a[i + j + k] = u - v;}}}
}
void idft(std::vector<Z> &a) {int n = a.size();std::reverse(a.begin() + 1, a.end());dft(a);Z inv = (1 - P) / n;for (int i = 0; i < n; i++) {a[i] *= inv;}
}
struct Poly {std::vector<Z> a;Poly() {}Poly(const std::vector<Z> &a) : a(a) {}Poly(const std::initializer_list<Z> &a) : a(a) {}int size() const {return a.size();}void resize(int n) {a.resize(n);}Z operator[](int idx) const {if (idx < size()) {return a[idx];} else {return 0;}}Z &operator[](int idx) {return a[idx];}Poly mulxk(int k) const {auto b = a;b.insert(b.begin(), k, 0);return Poly(b);}Poly modxk(int k) const {k = std::min(k, size());return Poly(std::vector<Z>(a.begin(), a.begin() + k));}Poly divxk(int k) const {if (size() <= k) {return Poly();}return Poly(std::vector<Z>(a.begin() + k, a.end()));}friend Poly operator+(const Poly &a, const Poly &b) {std::vector<Z> res(std::max(a.size(), b.size()));for (int i = 0; i < int(res.size()); i++) {res[i] = a[i] + b[i];}return Poly(res);}friend Poly operator-(const Poly &a, const Poly &b) {std::vector<Z> res(std::max(a.size(), b.size()));for (int i = 0; i < int(res.size()); i++) {res[i] = a[i] - b[i];}return Poly(res);}friend Poly operator*(Poly a, Poly b) {if (a.size() == 0 || b.size() == 0) {return Poly();}int sz = 1, tot = a.size() + b.size() - 1;while (sz < tot) {sz *= 2;}a.a.resize(sz);b.a.resize(sz);dft(a.a);dft(b.a);for (int i = 0; i < sz; ++i) {a.a[i] = a[i] * b[i];}idft(a.a);a.resize(tot);return a;}friend Poly operator*(Z a, Poly b) {for (int i = 0; i < int(b.size()); i++) {b[i] *= a;}return b;}friend Poly operator*(Poly a, Z b) {for (int i = 0; i < int(a.size()); i++) {a[i] *= b;}return a;}Poly &operator+=(Poly b) {return (*this) = (*this) + b;}Poly &operator-=(Poly b) {return (*this) = (*this) - b;}Poly &operator*=(Poly b) {return (*this) = (*this) * b;}Poly deriv() const {if (a.empty()) {return Poly();}std::vector<Z> res(size() - 1);for (int i = 0; i < size() - 1; ++i) {res[i] = (i + 1) * a[i + 1];}return Poly(res);}Poly integr() const {std::vector<Z> res(size() + 1);for (int i = 0; i < size(); ++i) {res[i + 1] = a[i] / (i + 1);}return Poly(res);}Poly inv(int m) const {Poly x{a[0].inv()};int k = 1;while (k < m) {k *= 2;x = (x * (Poly{2} - modxk(k) * x)).modxk(k);}return x.modxk(m);}Poly log(int m) const {return (deriv() * inv(m)).integr().modxk(m);}Poly exp(int m) const {Poly x{1};int k = 1;while (k < m) {k *= 2;x = (x * (Poly{1} - x.log(k) + modxk(k))).modxk(k);}return x.modxk(m);}Poly pow(int k, int m) const {int i = 0;while (i < size() && a[i].val() == 0) {i++;}if (i == size() || 1LL * i * k >= m) {return Poly(std::vector<Z>(m));}Z v = a[i];auto f = divxk(i) * v.inv();return (f.log(m - i * k) * k).exp(m - i * k).mulxk(i * k) * power(v, k);}Poly sqrt(int m) const {Poly x{1};int k = 1;while (k < m) {k *= 2;x = (x + (modxk(k) * x.inv(k)).modxk(k)) * ((P + 1) / 2);}return x.modxk(m);}Poly mulT(Poly b) const {if (b.size() == 0) {return Poly();}int n = b.size(); // eb + 1std::reverse(b.a.begin(), b.a.end());return ((*this) * b).divxk(n - 1); //保留系数(x ^ eb)及以上的}std::vector<Z> eval(std::vector<Z> x) const {if (size() == 0) {return std::vector<Z>(x.size(), 0);}const int n = std::max(int(x.size()), size());std::vector<Poly> q(4 * n);std::vector<Z> ans(x.size());x.resize(n);std::function<void(int, int, int)> build = [&](int p, int l, int r) {if (r - l == 1) {q[p] = Poly{1, -x[l]};} else {int m = (l + r) / 2;build(2 * p, l, m);build(2 * p + 1, m, r);q[p] = q[2 * p] * q[2 * p + 1];}};build(1, 0, n);std::function<void(int, int, int, const Poly &)> work = [&](int p, int l, int r, const Poly &num) {if (r - l == 1) {if (l < int(ans.size())) {ans[l] = num[0];}} else {int m = (l + r) / 2;work(2 * p, l, m, num.mulT(q[2 * p + 1]).modxk(m - l));work(2 * p + 1, m, r, num.mulT(q[2 * p]).modxk(r - m));}};work(1, 0, n, mulT(q[1].inv(n)));return ans;}
};void solve(){int n, m, k;cin >> n >> m >> k;vector<Z> a(k + 1);a[0] = 1;int R = min(n, k);for(int i = 1;i <= R;i++){a[i] = a[i - 1] * (n - i + 1) / i;}for(int i = 1;i <= R;i+=2){a[i] = 0;}for(int i = 0;i <= R;i+=2){a[i] *= 2;}auto c = Poly(a).pow(m,k + 1);cout << c[k] << "\n";
}

L-Misaka Mikoto’s Dynamic KMP Problem_2023牛客暑期多校训练营7 (nowcoder.com)

思路

注意到|s|>|t|时出现次数必定为0，所以该询问的$x_i\times y_i=0$。而当$|s|\le |t|$直接暴力跑KMP求出现次数和最长border即可。单轮复杂度O(t)。

代码

#include <bits/stdc++.h>
using namespace std;
using ll = long long;template<class T>
class KMP {int m;T p;public:vector<int> nxt;KMP() {}KMP(const T &_p) { init(_p); }void init(const T &_p) {m = _p.size() - 1;p = _p;nxt.assign(m + 1, 0);for (int i = 2;i <= m;i++) {nxt[i] = nxt[i - 1];while (nxt[i] && p[i] != p[nxt[i] + 1]) nxt[i] = nxt[nxt[i]];nxt[i] += p[i] == p[nxt[i] + 1];}}vector<int> find(const T &s) {int n = s.size() - 1;vector<int> pos;for (int i = 1, j = 0;i <= n;i++) {while (j && s[i] != p[j + 1]) j = nxt[j];j += s[i] == p[j + 1];if (j == m) {pos.push_back(i - j + 1);j = nxt[j];}}return pos;}vector<int> get_cycle_time() {vector<int> res;int pos = m;while (pos) {pos = nxt[pos];res.push_back(m - pos);}return res;}vector<int> get_cycle_loop() {vector<int> res;for (auto val : get_cycle_time())if (!(m % val)) res.push_back(val);return res;}int min_cycle_loop() { return get_cycle_loop()[0]; }void debug() {for (int i = 1;i <= m;i++)cout << nxt[i] << " \n"[i == m];}
};
/// KMP，前缀函数O(|P|)、查找O(|S|+|P|)、循环相关O(|P|)，维护字符串前缀函数int main() {std::ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);int n, q, b, p;cin >> n >> q >> b >> p;vector<int> s(n + 1);for (int i = 1;i <= n;i++) cin >> s[i];ll z = 0;int mul = 1, ans = 0;while (q--) {int op;cin >> op;if (op == 1) {ll x, c;cin >> x >> c;x = (x ^ z) % n + 1;c ^= z;s[x] = c;}else {int m;cin >> m;vector<int> t(m + 1);for (int i = 1;i <= m;i++) {ll val;cin >> val;t[i] = val ^ z;}mul = 1LL * mul * b % p;if (m < n) z = 0;else {KMP<vector<int>> kmp(s);z = 1LL * kmp.nxt[n] * kmp.find(t).size();}ans = (ans + z % p * mul % p) % p;}}cout << ans << '\n';return 0;
}

M-Writing Books_2023牛客暑期多校训练营7 (nowcoder.com)

思路

直接枚举位数。

代码

#include <bits/stdc++.h>
#define int long long
#define endl '\n'
#define IOS ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define fi first
#define sc secondusing namespace std;
const int INF = 0x3f3f3f3f3f3f3f3f;
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
typedef pair<int, int> PII;
int n;
int a[N];void solve() {cin >> n;int tmp = 1;int res = 0;int cnt = 1;while (n >= tmp) {int x = tmp;tmp *= 10;res += (min(n, tmp - 1) - x + 1) * cnt;cnt++;}cout << res << "\n";
}signed main() {IOS;int t = 1;cin >> t;for (int i = 1; i <= t; i++) {solve();}
}