[ECNU] 3314. 多项式展开 (1)
作者：互联网
https://acm.ecnu.edu.cn/problem/3314/
二项式展开得：
$$
\begin{aligned}
&\sum_{i=0}^{n}a_i(x+A)^i \\
=&\sum_{i=0}^{n}a_i\sum_{j=0}^{i}{i \choose j} x^jA^{i-j} \\
=&\sum_{j=0}^{n}x^j\sum_{i=j}^{n}{i \choose j}a_iA^{i-j} \\
=&\sum_{j=0}^{n}x^j\sum_{i=j}^{n}\frac{i!}{j!(i-j)!}a_iA^{i-j} \\
&b_i=\sum_{j=i}^{n}\frac{j!}{i!(j-i)!}a_jA^{j-i} \\
&i! \cdot A^i \cdot b_i=\sum_{j=i}^{n}\frac{j!}{(j-i)!}a_jA^{j} \\
& p_i=\sum_{j=i}^{n}q_jw_{j-i} \\
& p_i=\sum_{j=0}^{n-i}q_{j+i}w_j \\
& p'_i=p_{n-i} \\
& p'_{n-i}=\sum_{j=0}^{n-i}q_{j+i}w_j \\
& p'_i=\sum_{j=0}^{i}q_{j+n-i}w_j \\
& q'_{i}=q_{n-i},q_{j+n-i}=q'_{i-j} \\
& p'_i=\sum_{j=0}^{i}q'_{i-j}w_j
\end{aligned}
$$
只需要对$q',w$求一次卷积即可
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 const int mod = 998244353;
5 const int N = 2e6;
6 ll pw(ll a, ll b) {
7 ll res = 1;
8 while(b) {
9 if(b & 1) {
10 res = res * a % mod;
11 }
12 a = a * a % mod;
13 b >>= 1;
14 }
15 return res;
16 }
17 ll fac[N], invfac[N];
18 ll a[N], q[N], q_[N], w[N], b[N];
19 ll p_[N], p[N];
20 ll A, pwA[N], invpwA[N];
21 int n;
22
23 namespace NTT {
24 const int mod = 998244353, g[2] = { 3, (mod + 1) / 3 };
25 ll a[N], b[N], f[N];
26 int n, m;
27 int rev(int x, int n) {
28 int r = 0;
29 for(int i = 0 ; (1 << i) < n ; ++ i) {
30 r = (r << 1) | ((x >> i) & 1);
31 }
32 return r;
33 }
34 void ntt(ll *a, int ty, int n) {
35 for(int i = 0 ; i < n ; ++ i) {
36 f[rev(i, n)] = a[i];
37 }
38 for(int i = 2 ; i <= n ; i <<= 1) {
39 ll wn = pw(g[ty], (mod - 1) / i);
40 for(int j = 0 ; j < n ; j += i) {
41 ll w = 1;
42 for(int k = j ; k < j + i / 2 ; ++ k) {
43 ll u = f[k], v = f[k + i / 2] * w % mod;
44 f[k] = (u + v) % mod;
45 f[k + i / 2] = (u - v) % mod;
46 w = w * wn % mod;
47 }
48 }
49 }
50 for(int i = 0, inv = pw(n, mod - 2) ; i < n ; ++ i) {
51 a[i] = f[i];
52 if(ty) {
53 a[i] = a[i] * inv % mod;
54 }
55 }
56 }
57 void ntt() {
58 int len = 1;
59 while(len <= 2 * (n + m)) len <<= 1;
60 ntt(a, 0, len), ntt(b, 0, len);
61 for(int i = 0 ; i < len ; ++ i) {
62 a[i] = a[i] * b[i] % mod;
63 }
64 ntt(a, 1, len);
65 }
66 };
67
68 int main() {
69 scanf("%d", &n);
70 for(int i = 0 ; i <= n ; ++ i) {
71 scanf("%lld", &a[i]);
72 }
73 scanf("%lld", &A);
74
75 if(A == 0) {
76 for(int i = 0 ; i <= n ; ++ i) {
77 printf("%lld ", a[i]);
78 }
79 return 0;
80 }
81
82 pwA[0] = 1;
83 for(int i = 1 ; i <= n ; ++ i) {
84 pwA[i] = pwA[i - 1] * A % mod;
85 }
86 invpwA[n] = pw(pwA[n], mod - 2);
87 for(int i = n - 1 ; i >= 0 ; -- i) {
88 invpwA[i] = invpwA[i + 1] * A % mod;
89 }
90 fac[0] = 1;
91 for(int i = 1 ; i <= n ; ++ i) {
92 fac[i] = fac[i - 1] * i % mod;
93 }
94 invfac[n] = pw(fac[n], mod - 2);
95 for(int i = n - 1 ; i >= 0 ; -- i) {
96 invfac[i] = invfac[i + 1] * (i + 1) % mod;
97 }
98 for(int i = 0 ; i <= n ; ++ i) {
99 w[i] = invfac[i];
100 }
101 for(int i = 0 ; i <= n ; ++ i) {
102 q[i] = fac[i] * a[i] % mod * pw(A, i) % mod;
103 q_[n - i] = q[i];
104 }
105
106 NTT :: n = NTT :: m = n;
107 for(int i = 0 ; i <= n ; ++ i) {
108 NTT :: a[i] = q_[i];
109 NTT :: b[i] = w[i];
110 }
111 NTT :: ntt();
112 for(int i = 0 ; i <= n ; ++ i) {
113 p[i] = NTT :: a[n - i];
114 }
115 for(int i = 0 ; i <= n ; ++ i) {
116 b[i] = p[i] * invfac[i] % mod * pw(pw(A, i), mod - 2) % mod;
117 }
118
119 for(int i = 0 ; i <= n ; ++ i) {
120 printf("%lld ", (b[i] + mod) % mod);
121 }
122 }
View Code
标签：const,int,多项式,sum,3314,res,ECNU,ll,mod 来源： https://www.cnblogs.com/nekko/p/15378279.html