HDU 6061 - RXD and functions | 2017 Multi-University Training Contest 3
每次NTT都忘记初始化,真的是写一个小时,Debug两个小时- -
- /*
- HDU 6061 - RXD and functions [ NTT ] | 2017 Multi-University Training Contest 3
- 题意:
- 给定多项式 F(x) = ∑[0<=i<=n] f(i)*x^i
- 求多项式 G(x) = F(x-a)
- n <= 1e5
- 分析:
- 设 G(x) = ∑ g(i)*x^i
- 将 F(x-a) 按二项式定理展开后易得:
- g(x) = ∑[x<=y<=n] Comb(y,x) * f(y) * (-a)^(y-x)
- 打开组合数,移项:
- g(x)*x! = ∑[x<=y<=n] f(y)*y! * (-a)^(y-x) / (y-x)!
- 设 g'(x) = g(x)*x!
- p(x) = f(y)*y!
- q(x) = (-a)^x/x!
- 则 g'(x) = ∑[x<=y<=n] p(y) * q(y-x)
- = ∑[1<=y<=n-x] p(y+x) * q(y)
- 设 g''(x) = g'(n-x)
- 则 g''(x) = ∑[1<=y<=x] p(n-(x-y)) * q(y)
- 设 p'(x) = p(n-x)
- 则 g''(x) = ∑[1<=y<=x] p'(x-y) * q(y)
- 算出这个卷积后回带即可
- */
- #include <bits/stdc++.h>
- using namespace std;
- #define LL long long
- const int N = 1e5+5;
- const LL MOD = 998244353;
- namespace NTT {
- const int G = 3;
- const int NUM = 20;
- LL wn[20];
- LL mul(LL x, LL y) {
- return x*y% MOD;
- }
- LL PowMod(LL a, LL b) {
- LL res = 1;
- a %= MOD;
- while (b) {
- if (b&1) res = mul(res, a);
- a = mul(a, a);
- b >>= 1;
- }
- return res;
- }
- void Init() {
- for (int i = 0; i < NUM; i++)
- {
- int t = 1<<i;
- wn[i] = PowMod(G, (MOD-1)/t);
- }
- }
- void Change(LL a[], int len)
- {
- int i, j, k;
- for (i = 1, j = len/2; i < len-1; i++)
- {
- if (i < j) swap(a[i], a[j]);
- k = len/2;
- while (j >= k) {
- j -= k;
- k /= 2;
- }
- if (j < k) j += k;
- }
- }
- void NTT(LL a[], int len, int on)
- {
- Change(a, len);
- int id = 0;
- for (int h = 2; h <= len; h <<= 1)
- {
- id++;
- for (int j = 0; j < len; j += h)
- {
- LL w = 1;
- for (int k = j; k < j + h/2; k++)
- {
- LL u = a[k] % MOD;
- LL t = mul(a[k+h/2], w);
- a[k] = (u+t) % MOD;
- a[k+h/2] = ((u-t)% MOD + MOD ) % MOD;
- w = mul(w, wn[id]);
- }
- }
- }
- if (on == -1) {
- for (int i = 1; i < len/2; i++)
- swap(a[i], a[len-i]);
- LL inv = PowMod(len, MOD-2);
- for (int i = 0; i < len; i++)
- a[i] = mul(a[i], inv);
- }
- }
- void solve(LL a[], int n, LL b[], int m)
- {
- int len = 1;
- while (len < n*2 || len < m*2) len <<= 1;
- for (int i = n; i < len; i++) a[i] = 0;
- for (int i = m; i < len; i++) b[i] = 0;
- NTT(a, len, 1);
- NTT(b, len, 1);
- for (int i = 0; i < len; i++) a[i] = mul(a[i], b[i]);
- NTT(a, len, -1);
- }
- }
- LL f[N], p[N<<3], q[N<<3];
- LL a;
- int n, m;
- LL F[N], Finv[N], inv[N];
- void init(){
- inv[1] = 1;
- for (int i = 2; i < N; i++) {
- inv[i] = (MOD-MOD/i) * inv[MOD%i] % MOD;
- }
- F[0] = Finv[0] = 1;
- for (int i = 1; i < N; i++){
- F[i] = F[i-1] * i % MOD;
- Finv[i] = Finv[i-1] * inv[i] % MOD;
- }
- }
- void solve()
- {
- for (int i = 0; i <= n; i++)
- p[i] = F[i] * f[i] % MOD;//p(x)
- q[0] = 1;
- for (int i = 1; i <= n; i++)
- q[i] = q[i-1] * (MOD-a) % MOD * inv[i] % MOD;//q(x)
- for (int i = 0; i <= n/2; i++)
- swap(p[i], p[n-i]);//p'(x)
- NTT::solve(p, n+1, q, n+1);//g''(x)
- for (int i = 0; i <= n/2; i++)
- swap(p[i], p[n-i]);//g'(x)
- for (int i = 0; i <= n; i++)
- p[i] = p[i] * Finv[i] % MOD;//g(x)
- }
- int main()
- {
- init();
- NTT::Init();
- while (~scanf("%d", &n))
- {
- for (int i = 0; i <= n; i++) scanf("%lld", &f[i]);
- scanf("%d", &m);
- a = 0;
- for (int i = 1; i <= m; i++)
- {
- LL x; scanf("%lld", &x);
- a = (a+x) % MOD;
- }
- solve();
- for (int i = 0; i <= n; i++) printf("%lld ", p[i]);
- puts("");
- }
- }
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