HDU 6061 RXD and functions（NTT+卷积）

传送门

RXD and functions

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 532 Accepted Submission(s): 211


[align=left]Problem Description[/align] RXD has a polynomial function f(x), f(x)=∑ni=0cixi
RXD has a transformation of function Tr(f,a), it returns another function g, which has a property that g(x)=f(x−a).
Given a1,a2,a3,…,am, RXD generates a polynomial function sequence gi, in which g0=f and gi=Tr(gi−1,ai)
RXD wants you to find gm, in the form of ∑mi=0bixi
You need to output bi module 998244353.
n≤105
[align=left]Input[/align] There are several test cases, please keep reading until EOF.
For each test case, the first line consists of 1 integer n, which means degF.
The next line consists of n+1 intergers ci,0≤ci≤998244353, which means the coefficient of the polynomial.
The next line contains an integer m, which means the length of a.
The next line contains m integers, the i - th integer is ai.
There are 11 test cases.

0≤ai≤998244353

∑m≤105
[align=left]Output[/align] For each test case, output an polynomial with degree n, which means the answer.
[align=left]Sample Input[/align]

2

0 0 1

1

1

[align=left]Sample Output[/align]

1 998244351 1

Hint

(x−1)2=x2−2x+1

题目大意：

已知多项式 f(x)=∑ni=0cixi 求 f(x−∑ai)

解题思路：

令 −t=∑ai

f(x)=∑ni=0ci(x+t)i

将 (x+t)i用二项式定理展开后得到：

f(x)将组合数展开：将i!提到前面：=∑i=0nci∑j=0ixjCjiti−j=∑i=0nxi∑j=incj∗Cijtj−i=∑i=0nxi∑j=incj∗j!i!∗(j−i)!tj−i=∑i=0nxii!∑j=incj∗j!tj−i(j−i)!

我们将其系数提出来：f′(x)=∑j=incj∗j!tj−i(j−i)!

令 bi=cn−i∗(n−i)!，则有：

f′(x)=∑j=inbn−jtj−i(j−i)!

通过观察发现： (n−j)+(j−i)=n−i

现在设：h=n−i,则有：

f′(x)=∑j=0hbh−jtjj!

这是个卷积形式，因为是模意义下的，我们可以根据 NTT 求出f′(x)的系数，但是在真正的 f(x) 里，f(i)=f′(n−i)i!。

我们需要预处理阶乘逆元，以及阶乘，然后通过 NTT 计算 f′(x)，最后计算 f(x) 系数就OK了。

代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 3e5+5;
const LL MOD = 998244353;
const double eps = 1e-8;
const double PI = acos(-1.0);
LL c[100005], fac[100005], Inv[100005], ans[MAXN];
void Init(){
fac[0] = Inv[0] = fac[1] = Inv[1] = 1;
for(int i=2; i<100005; i++) fac[i] = fac[i-1] * i % MOD;
for(int i=2; i<100005; i++) Inv[i] = (MOD - MOD / i) * Inv[MOD % i] % MOD;
for(int i=2; i<100005; i++) Inv[i] = Inv[i] * Inv[i-1] % MOD;
}
///const LL P = (479 << 21) + 1;//费马素数
const LL P = MOD;
const LL G = 3;//原根
const LL NUM = 20;

LL  wn[NUM];

LL Pow(LL a, LL b, LL m){
LL ans = 1;
a %= m;
while(b){
if(b & 1) ans = ans*a%m;
b>>=1;
a = a*a%m;
}
return ans;
}
void GetWn()
{
for(int i = 0; i < NUM; i++)
{
LL t = 1LL << i;
wn[i] = Pow(G, (P - 1) / t, P);
}
}
void change(LL * y, LL len) {
LL i, j, k;
for (i = 1, j = len / 2; i < len - 1; i++) {
if (i < j) swap(y[i], y[j]);
k = len / 2;
while (j >= k) {
j -= k;
k /= 2;
}
if (j < k) j += k;
}
}

void ntt(LL *y, LL len, LL on) {
change(y, len);
LL id = 0;
for (LL h = 2; h <= len; h <<= 1) {
id++;
for (LL j = 0; j < len; j += h) {
LL w = 1;
for (LL k = j; k < j + h / 2; k++) {
LL u = y[k] % P;
LL t = w * y[k + h / 2] % P;
y[k] = (u + t) % P;
y[k + h / 2] = (u - t + P) % P;
w = w * wn[id] % P;
}
}
}
if(on == -1)
{
for(LL i = 1; i < len / 2; i++)
swap(y[i], y[len - i]);
LL inv = Pow(len, P - 2, P);
for(LL i = 0; i < len; i++)
y[i] = y[i] * inv % P;
}
}
LL a[MAXN], b[MAXN];
int main()
{
///freopen("in.txt", "r", stdin);
Init();
GetWn();
int n;
while(~scanf("%d", &n)){
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
n++;
for(int i=0; i<n; i++) scanf("%lld", &c[i]);
int m; scanf("%d", &m);
LL x, sum = 0;
while(m--) scanf("%lld", &x), sum = (sum + x) % MOD;
sum = (-sum + MOD) % MOD;
if(sum == 0){
for(int i=0; i<n; i++) printf("%lld ", c[i]);
puts("");
continue;
}
LL len = 1;
while(len < (2*n)) len<<=1LL;
LL s = 1;
for(int i=0; i<n; i++){
a[i] = c[n-1-i] * fac[n-1-i] % MOD;
b[i] = s * Inv[i] % MOD;
s = s * sum % MOD;
}
ntt(a, len, 1), ntt(b, len, 1);
for(int i=0; i<len; i++) a[i] = a[i] * b[i] % MOD;
ntt(a, len, -1);
for(int i=0; i<n; i++) printf("%lld ",a[n-1-i]*Inv[i]%MOD);
puts("");
}
return 0;
}