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  1. #include <bits/stdc++.h>
  2.  
  3. using namespace std;
  4.  
  5. const int md = 998244353;
  6.  
  7. int add(int x, int y) {
  8.   x += y;
  9.   if (x >= md) {
  10.     x -= md;
  11.   }
  12.   return x;
  13. }
  14.  
  15. int sub(int x, int y) {
  16.   x -= y;
  17.   if (x < 0) {
  18.     x += md;
  19.   }
  20.   return x;
  21. }
  22.  
  23. int mul(int x, int y) {
  24.   return (long long)x * y % md;
  25. }
  26.  
  27. int power(int x, int y) {
  28.   int result = 1;
  29.   for (; y; y >>= 1, x = mul(x, x)) {
  30.     if (y & 1) {
  31.       result = mul(result, x);
  32.     }
  33.   }
  34.   return result;
  35. }
  36.  
  37. namespace ntt {
  38. int base = 1, root = -1, max_base = -1;
  39. vector<int> roots = {0, 1}, rev = {0, 1};
  40.  
  41. void init() {
  42.   int temp = md - 1;
  43.   max_base = 0;
  44.   while (!(temp & 1)) {
  45.     temp >>= 1;
  46.     ++max_base;
  47.   }
  48.   root = 2;
  49.   while (true) {
  50.     if (power(root, 1 << max_base) == 1 && power(root, 1 << max_base - 1) != 1) {
  51.       break;
  52.     }
  53.     ++root;
  54.   }
  55. }
  56.  
  57. void ensure_base(int nbase) {
  58.   if (max_base == -1) {
  59.     init();
  60.   }
  61.   if (nbase <= base) {
  62.     return;
  63.   }
  64.   assert(nbase <= max_base);
  65.   rev.resize(1 << nbase);
  66.   for (int i = 0; i < (1 << nbase); ++i) {
  67.     rev[i] = rev[i >> 1] >> 1 | ((i & 1) << nbase - 1);
  68.   }
  69.   roots.resize(1 << nbase);
  70.   while (base < nbase) {
  71.     int z = power(root, 1 << max_base - 1 - base);
  72.     for (int i = 1 << base - 1; i < 1 << base; ++i) {
  73.       roots[i << 1] = roots[i];
  74.       roots[i << 1 | 1] = mul(roots[i], z);
  75.     }
  76.     ++base;
  77.   }
  78. }
  79.  
  80. void dft(vector<int> &a) {
  81.   int n = a.size(), zeros = __builtin_ctz(n);
  82.   ensure_base(zeros);
  83.   int shift = base - zeros;
  84.   for (int i = 0; i < n; ++i) {
  85.     if (i < rev[i] >> shift) {
  86.       swap(a[i], a[rev[i] >> shift]);
  87.     }
  88.   }
  89.   for (int i = 1; i < n; i <<= 1) {
  90.     for (int j = 0; j < n; j += i << 1) {
  91.       for (int k = 0; k < i; ++k) {
  92.         int x = a[j + k], y = mul(a[j + k + i], roots[i + k]);
  93.         a[j + k] = add(x, y);
  94.         a[j + k + i] = sub(x, y);
  95.       }
  96.     }
  97.   }
  98. }
  99.  
  100. vector<int> padd(vector<int> a, vector<int> b) {
  101.   if (a.size() < b.size()) {
  102.     a.resize(b.size());
  103.   }
  104.   for (int i = 0; i < b.size(); ++i) {
  105.     a[i] = add(a[i], b[i]);
  106.   }
  107.   return a;
  108. }
  109.  
  110. vector<int> psub(vector<int> a, vector<int> b) {
  111.   if (a.size() < b.size()) {
  112.     a.resize(b.size());
  113.   }
  114.   for (int i = 0; i < b.size(); ++i) {
  115.     a[i] = sub(a[i], b[i]);
  116.   }
  117.   return a;
  118. }
  119.  
  120. vector<int> pmul(vector<int> a, vector<int> b, bool equal = false) {
  121.   int need = a.size() + b.size() - 1, nbase = 0;
  122.   while (1 << nbase < need) {
  123.     ++nbase;
  124.   }
  125.   ensure_base(nbase);
  126.   int size = 1 << nbase;
  127.   a.resize(size);
  128.   b.resize(size);
  129.   dft(a);
  130.   if (equal) {
  131.     b = a;
  132.   } else {
  133.     dft(b);
  134.   }
  135.   int inv = power(size, md - 2);
  136.   for (int i = 0; i < size; ++i) {
  137.     a[i] = mul(mul(a[i], b[i]), inv);
  138.   }
  139.   reverse(a.begin() + 1, a.end());
  140.   dft(a);
  141.   a.resize(need);
  142.   return a;
  143. }
  144.  
  145. vector<int> psqr(vector<int> a) {
  146.   return pmul(a, a, true);
  147. }
  148.  
  149. vector<int> pinv(vector<int> a) {
  150.   int n = a.size(), m = n + 1 >> 1;
  151.   if (n == 1) {
  152.     return vector<int> (1, power(a[0], md - 2));
  153.   } else {
  154.     vector<int> b = pinv(vector<int> (a.begin(), a.begin() + m));
  155.     int need = n << 1, nbase = 0;
  156.     while (1 << nbase < need) {
  157.       ++nbase;
  158.     }
  159.     ensure_base(nbase);
  160.     int size = 1 << nbase;
  161.     a.resize(size);
  162.     b.resize(size);
  163.     dft(a);
  164.     dft(b);
  165.     int inv = power(size, md - 2);
  166.     for (int i = 0; i < size; ++i) {
  167.       a[i] = mul(mul(sub(2, mul(a[i], b[i])), b[i]), inv);
  168.     }
  169.     reverse(a.begin() + 1, a.end());
  170.     dft(a);
  171.     a.resize(n);
  172.     return a;
  173.   }
  174. }
  175.  
  176. vector<int> pdiv(vector<int> a, vector<int> b) {
  177.   int n = a.size(), m = b.size();
  178.   if (n < m) {
  179.     return vector<int> ();
  180.   }
  181.   reverse(a.begin(), a.end());
  182.   reverse(b.begin(), b.end());
  183.   b.resize(n - m + 1);
  184.   b = pmul(a, pinv(b));
  185.   b.erase(b.begin() + n - m + 1, b.end());
  186.   reverse(b.begin(), b.end());
  187.   return b;
  188. }
  189.  
  190. vector<int> pmod(vector<int> a, vector<int> b) {
  191.   int n = a.size(), m = b.size();
  192.   if (n < m) {
  193.     return a;
  194.   }
  195.   vector<int> c = pmul(pdiv(a, b), b);
  196.   for (int i = 0; i < m - 1; ++i) {
  197.     c[i] = sub(a[i], c[i]);
  198.   }
  199.   c.resize(m - 1);
  200.   return c;
  201. }
  202. }
  203.  
  204. using ntt::padd;
  205. using ntt::psub;
  206. using ntt::pmul;
  207. using ntt::psqr;
  208. using ntt::pinv;
  209. using ntt::pdiv;
  210. using ntt::pmod;
  211.  
  212. int main() {
  213. #ifdef wxh010910
  214.   freopen("input.txt", "r", stdin);
  215. #endif
  216.   int n;
  217.   scanf("%d", &n);
  218.   vector<int> a(n), b(n);
  219.   for (int i = 0; i < n; ++i) {
  220.     scanf("%d", &a[i]);
  221.   }
  222.   for (int i = 0; i < n; ++i) {
  223.     scanf("%d", &b[i]);
  224.   }
  225.  
  226.   function<vector<int>(int, int)> multiply_all = [&](int l, int r) {
  227.     if (l == r) {
  228.       return vector<int> {1, sub(0, a[l])};
  229.     } else {
  230.       int mid = l + r >> 1;
  231.       return pmul(multiply_all(l, mid), multiply_all(mid + 1, r));
  232.     }
  233.   };
  234.  
  235.   vector<vector<int>> segtree((n << 1) - 1);
  236.  
  237.   function<void(int, int, int)> build = [&](int x, int l, int r) {
  238.     if (l == r) {
  239.       segtree[x] = vector<int> {sub(0, power(a[l], md - 2)), 1};
  240.     } else {
  241.       int y = l + r >> 1, z = x + (y - l + 1 << 1);
  242.       build(x + 1, l, y);
  243.       build(z, y + 1, r);
  244.       segtree[x] = pmul(segtree[x + 1], segtree[z]);
  245.     }
  246.   };
  247.  
  248.   function<void(int, int, int, vector<int>, vector<int>&)> evaluate = [&](int x, int l, int r, vector<int> f, vector<int> &answer) {
  249.     f = pmod(f, segtree[x]);
  250.     if (l == r) {
  251.       answer[l] = f[0];
  252.     } else {
  253.       int y = l + r >> 1, z = x + (y - l + 1 << 1);
  254.       evaluate(x + 1, l, y, f, answer);
  255.       evaluate(z, y + 1, r, f, answer);
  256.     }
  257.   };
  258.  
  259.   vector<int> f = pmul(b, multiply_all(0, n - 1)), p(n), q(n);
  260.   f.resize(n);
  261.   build(0, 0, n - 1);
  262.   evaluate(0, 0, n - 1, f, p);
  263.   int coef = 1;
  264.   for (int i = 0; i < n; ++i) {
  265.     coef = mul(coef, sub(0, a[i]));
  266.   }
  267.   for (int i = 0; i < n; ++i) {
  268.     f[i] = mul(mul(n - i, segtree[0][i]), coef);
  269.   }
  270.   evaluate(0, 0, n - 1, f, q);
  271.  
  272.   for (int i = 0; i < n; ++i) {
  273.     printf("%d%c", mul(p[i], power(q[i], md - 2)), i == n - 1 ? '\n' : ' ');
  274.   }
  275.  
  276.   return 0;
  277. }
  278.  
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https://ideone.com/H7OmLn
language:
C++14 (gcc 8.3)
created:
5 years ago
visibility:
public

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