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
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 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
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
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 rev;
std::vector roots{0, 1};
void dft(std::vector &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 &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 a;
Poly() {}
Poly(const std::vector &a) : a(a) {}
Poly(const std::initializer_list &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(a.begin(), a.begin() + k));
}
Poly divxk(int k) const {
if (size() <= k) {
return Poly();
}
return Poly(std::vector(a.begin() + k, a.end()));
}
friend Poly operator+(const Poly &a, const Poly &b) {
std::vector 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 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 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 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(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 + 1
std::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 eval(std::vector x) const {
if (size() == 0) {
return std::vector(x.size(), 0);
}
const int n = std::max(int(x.size()), size());
std::vector q(4 * n);
std::vector ans(x.size());
x.resize(n);
std::function 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 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 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 × y i = 0 x_i×y_i=0 xi×yi=0。而当 ∣ s ∣ ≤ ∣ t ∣ |s| \leq |t| ∣s∣≤∣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
using namespace std;
using ll = long long;
template
class KMP {
int m;
T p;
public:
vector 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 find(const T &s) {
int n = s.size() - 1;
vector 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 get_cycle_time() {
vector res;
int pos = m;
while (pos) {
pos = nxt[pos];
res.push_back(m - pos);
}
return res;
}
vector get_cycle_loop() {
vector 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 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 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> 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
#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 second
using namespace std;
const int INF = 0x3f3f3f3f3f3f3f3f;
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
typedef pair 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();
}
}