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willy108

Rock Paper Scissors Hard

Aug 6th, 2022
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C++ 6.41 KB | None | 0 0
  1. #include <bits/stdc++.h>
  2. using namespace std;
  3.  
  4. const int N = 100000;
  5. namespace fft {
  6.   #define rep(i, a, b) for(int i = a; i < (b); ++i)
  7. #define trav(a, x) for(auto& a : x)
  8. #define all(x) x.begin(), x.end()
  9. #define sz(x) (int)(x).size()
  10. typedef long long ll;
  11. typedef pair<int, int> pii;
  12. typedef vector<int> vi;
  13. #define FFT 1
  14. #if FFT
  15. // FFT
  16. using dbl = double;
  17. struct num { /// start-hash
  18.   dbl x, y;
  19.   num(dbl x_ = 0, dbl y_ = 0) : x(x_), y(y_) { }
  20. };
  21. inline num operator+(num a, num b) { return num(a.x + b.x, a.y + b.y); }
  22. inline num operator-(num a, num b) { return num(a.x - b.x, a.y - b.y); }
  23. inline num operator*(num a, num b) { return num(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x); }
  24. inline num conj(num a) { return num(a.x, -a.y); }
  25. inline num inv(num a) { dbl n = (a.x*a.x+a.y*a.y); return num(a.x/n,-a.y/n); }
  26. /// end-hash
  27. #else
  28. // NTT
  29. const int mod = 998244353, g = 3;
  30. // For p < 2^30 there is also (5 << 25, 3), (7 << 26, 3),
  31. // (479 << 21, 3) and (483 << 21, 5). Last two are > 10^9.
  32. struct num { /// start-hash
  33.   int v;
  34.   num(ll v_ = 0) : v(int(v_ % mod)) { if (v<0) v+=mod; }
  35.   explicit operator int() const { return v; }
  36. };
  37. inline num operator+(num a,num b){return num(a.v+b.v);}
  38. inline num operator-(num a,num b){return num(a.v+mod-b.v);}
  39. inline num operator*(num a,num b){return num(1ll*a.v*b.v);}
  40. inline num pow(num a, int b) {
  41.   num r = 1;
  42.   do{if(b&1)r=r*a;a=a*a;}while(b>>=1);
  43.   return r;
  44. }
  45. inline num inv(num a) { return pow(a, mod-2); }
  46. /// end-hash
  47. #endif
  48.  
  49. using vn = vector<num>;
  50. vi rev({0, 1});
  51. vn rt(2, num(1)), fa, fb;
  52.  
  53. const double PI = acos(-1);
  54.  
  55. inline void init(int n) { /// start-hash
  56.   if (n <= sz(rt)) return;
  57.   rev.resize(n);
  58.   rep(i,0,n) rev[i] = (rev[i>>1] | ((i&1)*n)) >> 1;
  59.   rt.reserve(n);
  60.   for (int k = sz(rt); k < n; k *= 2) {
  61.     rt.resize(2*k);
  62. #if FFT
  63.     double a=PI/k; num z(cos(a),sin(a)); // FFT
  64. #else
  65.     num z = pow(num(g), (mod-1)/(2*k)); // NTT
  66. #endif
  67.     rep(i,k/2,k) rt[2*i] = rt[i], rt[2*i+1] = rt[i]*z;
  68.   }
  69. } /// end-hash
  70.  
  71. inline void fft(vector<num> &a, int n) { /// start-hash
  72.   init(n);
  73.   int s = __builtin_ctz(sz(rev)/n);
  74.   rep(i,0,n) if (i < rev[i]>>s) swap(a[i], a[rev[i]>>s]);
  75.   for (int k = 1; k < n; k *= 2)
  76.     for (int i = 0; i < n; i += 2 * k) rep(j,0,k) {
  77.       num t = rt[j+k] * a[i+j+k];
  78.       a[i+j+k] = a[i+j] - t;
  79.       a[i+j] = a[i+j] + t;
  80.     }
  81. } /// end-hash
  82.  
  83. // Complex/NTT
  84. vn multiply(vn a, vn b) { /// start-hash
  85.   int s = sz(a) + sz(b) - 1;
  86.   if (s <= 0) return {};
  87.   int L = s > 1 ? 32 - __builtin_clz(s-1) : 0, n = 1 << L;
  88.   a.resize(n), b.resize(n);
  89.   fft(a, n);
  90.   fft(b, n);
  91.   num d = inv(num(n));
  92.   rep(i,0,n) a[i] = a[i] * b[i] * d;
  93.   reverse(a.begin()+1, a.end());
  94.   fft(a, n);
  95.   a.resize(s);
  96.   return a;
  97. } /// end-hash
  98.  
  99. // Complex/NTT power-series inverse
  100. // Doubles b as b[:n] = (2 - a[:n] * b[:n/2]) * b[:n/2]
  101. vn inverse(const vn& a) { /// start-hash
  102.   if (a.empty()) return {};
  103.   vn b({inv(a[0])});
  104.   b.reserve(2*a.size());
  105.   while (sz(b) < sz(a)) {
  106.     int n = 2*sz(b);
  107.     b.resize(2*n, 0);
  108.     if (sz(fa) < 2*n) fa.resize(2*n);
  109.     fill(fa.begin(), fa.begin()+2*n, 0);
  110.     copy(a.begin(), a.begin()+min(n,sz(a)), fa.begin());
  111.     fft(b, 2*n);
  112.     fft(fa, 2*n);
  113.     num d = inv(num(2*n));
  114.     rep(i, 0, 2*n) b[i] = b[i] * (2 - fa[i] * b[i]) * d;
  115.     reverse(b.begin()+1, b.end());
  116.     fft(b, 2*n);
  117.     b.resize(n);
  118.   }
  119.   b.resize(a.size());
  120.   return b;
  121. } /// end-hash
  122.  
  123. #if FFT
  124. // Double multiply (num = complex)
  125. using vd = vector<double>;
  126. vd multiply(const vd& a, const vd& b) { /// start-hash
  127.   int s = sz(a) + sz(b) - 1;
  128.   if (s <= 0) return {};
  129.   int L = s > 1 ? 32 - __builtin_clz(s-1) : 0, n = 1 << L;
  130.   if (sz(fa) < n) fa.resize(n);
  131.   if (sz(fb) < n) fb.resize(n);
  132.  
  133.   fill(fa.begin(), fa.begin() + n, 0);
  134.   rep(i,0,sz(a)) fa[i].x = a[i];
  135.   rep(i,0,sz(b)) fa[i].y = b[i];
  136.   fft(fa, n);
  137.   trav(x, fa) x = x * x;
  138.   rep(i,0,n) fb[i] = fa[(n-i)&(n-1)] - conj(fa[i]);
  139.   fft(fb, n);
  140.   vd r(s);
  141.   rep(i,0,s) r[i] = fb[i].y / (4*n);
  142.   return r;
  143. } /// end-hash
  144.  
  145. // Integer multiply mod m (num = complex) /// start-hash
  146. vi multiply_mod(const vi& a, const vi& b, int m) {
  147.   int s = sz(a) + sz(b) - 1;
  148.   if (s <= 0) return {};
  149.   int L = s > 1 ? 32 - __builtin_clz(s-1) : 0, n = 1 << L;
  150.   if (sz(fa) < n) fa.resize(n);
  151.   if (sz(fb) < n) fb.resize(n);
  152.  
  153.   rep(i,0,sz(a)) fa[i] = num(a[i] & ((1<<15)-1), a[i] >> 15);
  154.   fill(fa.begin()+sz(a), fa.begin() + n, 0);
  155.   rep(i,0,sz(b)) fb[i] = num(b[i] & ((1<<15)-1), b[i] >> 15);
  156.   fill(fb.begin()+sz(b), fb.begin() + n, 0);
  157.  
  158.   fft(fa, n);
  159.   fft(fb, n);
  160.   double r0 = 0.5 / n; // 1/2n
  161.   rep(i,0,n/2+1) {
  162.     int j = (n-i)&(n-1);
  163.     num g0 = (fb[i] + conj(fb[j])) * r0;
  164.     num g1 = (fb[i] - conj(fb[j])) * r0;
  165.     swap(g1.x, g1.y); g1.y *= -1;
  166.     if (j != i) {
  167.       swap(fa[j], fa[i]);
  168.       fb[j] = fa[j] * g1;
  169.       fa[j] = fa[j] * g0;
  170.     }
  171.     fb[i] = fa[i] * conj(g1);
  172.     fa[i] = fa[i] * conj(g0);
  173.   }
  174.   fft(fa, n);
  175.   fft(fb, n);
  176.   vi r(s);
  177.   rep(i,0,s) r[i] = int((ll(fa[i].x+0.5)
  178.         + (ll(fa[i].y+0.5) % m << 15)
  179.         + (ll(fb[i].x+0.5) % m << 15)
  180.         + (ll(fb[i].y+0.5) % m << 30)) % m);
  181.   return r;
  182. } /// end-hash
  183. #endif
  184.  
  185. } // namespace fft
  186.  
  187. using namespace fft;
  188.  
  189. long long power(long long a, long long k, const int& md) {
  190.   long long p = 1;
  191.   while (k > 0) {
  192.     if (k & 1)
  193.       p = p * a % md;
  194.     a = a * a % md;
  195.     k >>= 1;
  196.   }
  197.   return p;
  198. }
  199. long long inv(long long a, const int& md) {
  200.   return a == 1 ? 1 : inv(a - md % a, md) * (md / a + 1) % md;
  201. }
  202.  
  203. int n, md;
  204. long long ff[N + 1], gg[N + 1], dp[N + 1];
  205.  
  206. void dnc(int l, int r) {
  207.   if (l == r)
  208.     dp[l] = (power(3, l - 1, md) + ff[l] * dp[l] % md) * inv(power(2, l, md) - 2 + md, md) % md;
  209.   else {
  210.     int m = (l + r) / 2;
  211.     dnc(l, m);
  212.     vector<int> a(m - l + 1), b(r - l + 1);
  213.     for (int i = l; i <= m; i++)
  214.       a[i - l] = dp[i] * gg[i] % md;
  215.     for (int i = 1; i <= r - l; i++)
  216.       b[i] = gg[i];
  217.     vector<int> c = multiply_mod(a, b, md);
  218.     for (int i = m + 1; i <= r; i++)
  219.       dp[i] = (dp[i] + (int) c[i - l]) % md;
  220.     dnc(m + 1, r);
  221.   }
  222. }
  223.  
  224. int main() {
  225.   ios::sync_with_stdio(false);
  226.   cin.tie(NULL);
  227.   cin >> n >> md;
  228.   ff[0] = 1;
  229.   for (int i = 1; i <= n; i++)
  230.     ff[i] = ff[i - 1] * i % md;
  231.   gg[n] = inv(ff[n], md);
  232.   for (int i = n; i >= 1; i--)
  233.     gg[i - 1] = gg[i] * i % md;
  234.   dnc(2, n);
  235.   for (int i = 1; i <= n; i++)
  236.     cout << dp[i] << ' ';
  237.   cout << '\n';
  238.   return 0;
  239. }
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