A Dance of Fire and Ice

你有一个数 \(val\),初始为 \(1\)。然后给你 \(n\) 次操作,第 \(i\) 次操作有参数 \(opt_{i}\)\(a_{i}\) ,其中 \(opt_{i} = 0/1\)

\(opt_{i} = 0\) 表示将 \(val\) 改为 \(a_{i}\)\(opt_{i} = 1\) 表示将 \(val\) 变为 \(val \times a_{i} \bmod p\)

现在给出 \(p\)\(n\) 以及 \(n\) 次操作,保证 \(p\) 为质数,你可以从中选出任意操作并以任意顺序进行所有操作,问最终 \(val\) 有多少种可能的取值。


乘法做背包是不好用 bitset 优化的,乘法转对数就可以是加法了,具体用原根。

#include <bits/stdc++.h>

using namespace std;

#define IL inline
#define vec vector
#define bg begin
#define eb emplace_back
#define emp emplace
#define fi first
#define se second
#define mkp make_pair
#define lb lower_bound
#define ub upper_bound
using ubt = long long;
using uubt = unsigned long long;
using dub = double;
using pii = pair<int, int>;

IL void ckx(ubt &x, const ubt &y) { (x < y) && (x = y); }
IL void ckm(ubt &x, const ubt &y) { (x > y) && (x = y); }

template <typename T = int>
IL T _R() {
  T s = 0, w = 1;
  char c = getchar();
  while (!isdigit(c)) w = c == '-' ? -1 : 1, c = getchar();
  while (isdigit(c)) s = s * 10 + c - 48, c = getchar();
  return s * w;
}

const dub inf = 1e18;

const int V = 2e5;
const int maxV = V + 1;

int mod, n;
int c[maxV];

bitset<V> f;

int v[maxV];

struct Qpow {
  static const int S = 5, K = (20 - 1) / S + 1, L = 1 << S;
  int r[K][L];
  IL void init(int b) {
    for (int i = 0; i < K; ++i) {
      r[i][0] = 1;
      for (int j = 1; j < L; ++j)
        r[i][j] = 1ll * r[i][j - 1] * b % mod;
      b = 1ll * b * r[i][L - 1] % mod;
    }
  }
  IL int operator () (int x) {
    int res = 1;
    for (int i = 0; i < K; ++i) {
      res = 1ll * res * r[i][x & (L - 1)] % mod;
      x >>= S;
    }
    return res;
  }
} qpow;

IL vec<int> pfac(int n) {
  vec<int> pf;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) {
      pf.eb(i);
      while (n % i == 0) n /= i;
    }
  if (n > 1) pf.eb(n);
  return pf;
}
int G(int n) {
  int phi = n - 1;
  auto pf = pfac(phi);
  for (int i = 1; i <= n; i++) {
    if (__gcd(i, n) ^ 1) continue;
    qpow.init(i);
    bool flag = true;
    for (auto j : pf)
      if (qpow(phi / j) == 1) {
        flag = false;
        break;
      }
    if (flag) return i;
  }
  assert(false);
  return -1;
}

int main() {
  freopen("dance.in", "r", stdin);
  freopen("dance.out", "w", stdout);

  mod = _R(), n = _R();

  int root = G(mod);
  qpow.init(root);
  for (int i = 0; i < mod - 1; i++)
    v[qpow(i)] = i;

  assert(v[1] == 0);

  int ans = 0;
  f[v[1]] = true;
  for (int i = 1; i <= n; i++) {
    int op = _R(), x = _R() % mod;
    if (!x) ans += !ans;
    else if (!op) f[v[x]] = true;
    else c[v[x]]++;
  }
  for (int i = 0; i < mod; i++)
    while (c[i]--) {
      auto h = f | (f << i) | (f >> (mod - 1 - i));
      if (f == h) break; // is important
      f = h;
    }
  for (int i = 0; i < mod - 1; i++) ans += f[i];
  printf("%d\n", ans);
}
posted @ 2025-01-15 16:16  ccxswl  阅读(78)  评论(10)    收藏  举报