[HDU] 6061 RXD and functions [NTT]

URL： http://acm.split.hdu.edu.cn/showproblem.php?pid=6061

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 524288/524288

K (Java/Others) Total Submission(s): 702 Accepted Submission(s): 282

Problem Description

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

Input

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

Output

For each test case, output an polynomial with degree n, which means the answer.

Sample Input

2

0 0 1

1

1

Sample Output

1 998244351 1

Hint

(x−1)2=x2−2x+1

题解

待求表达式可以拆开然后化成 a(n)=∑ni=0h(i)g(i) 的形式

注意到只要令 h′(x)=h(n−x) 就可到一个卷积式 (h⊗g)(n)=∑ni=0h′(n−i)g(i)

然后NTT

#include<stdio.h>

typedef long long LL;
const int MAXN = 1 << 18;
const int p    = (119 << 23) + 1;
const int g    = 3;
const int NUM  = 23;
int C[MAXN], A[MAXN], B[MAXN], fac[MAXN], wn[MAXN], n, m, len, S;

void swap(int &x, int &y) {
int t = x;
x = y;
y = t;
}

int mod_pow(int a, int k) {
int res = 1;
while(k) {
if(k&1) res = 1LL * res * a % p;
a = 1LL * a * a % p;
k >>= 1;
}
return res;
}

void init() {
fac[0] = 1;
for(int i = 1; i < MAXN; ++i) {
fac[i] = 1LL * fac[i - 1] * i % p;
}
for(int i = 0; i < NUM; ++i) {
int t = 1 << i;
wn[i] = mod_pow(g, (p - 1) / t);
}
}

void rader(int y[]) {
for(int i = 1, j = len >> 1, k; i < len - 1; ++ i) {
if(i < j) swap(y[i], y[j]);
k = len >> 1;
while(j >= k) {
j -= k;
k >>= 1;
}
if(j < k) j += k;
}
}

void NTT(int y[], int op) {
rader(y);
int id = 0;
for(int h = 2; h <= len; h <<= 1) {
++id;
for(int i = 0; i < len; i += h) {
int w = 1;
for(int j = i; j < i + h / 2; ++j) {
int u = y[j];
int t = 1LL * w * y[j + h / 2] % p;
y[j] = (u + t) % p;
y[j + h / 2] = (u - t) % p;
w = 1LL * w * wn[id] % p;
}
}
}
if(op == -1) {
for(int i = 1; i < len / 2; ++i) {
swap(y[i], y[len - i]);
}
int inv = mod_pow(len, p - 2);
for(int i = 0; i < len; ++i) {
y[i] = 1LL * y[i] * inv % p;
y[i] += p;
y[i] %= p;
}
}
}

void slove() {
for(len = 1; len < (n << 1); len <<= 1) ;
//  printf("len: %d\n", len);
for(int i = 0; i < len; ++i) {
A[i] = i > n ? 0 : 1LL * C[n - i] * fac[n - i] % p;
B[i] = i > n ? 0 : 1LL * mod_pow(S, i) * mod_pow(fac[i], p - 2) % p;
//      printf("#%d %d\n", C[i], B[i]);
}
NTT(A, 1);
NTT(B, 1);
for(int i = 0; i < len; ++i) {
B[i] = 1LL * A[i] * B[i] % p;
}
NTT(B, -1);
int top = len - 1;
while(top && !B[top]) --top;
for(int i = 0; i <= n; ++i) {
int ans = 1LL * B[n - i] * mod_pow(fac[i], p - 2) % p;
printf(i == n ? "%d \n" : "%d ", ans);
}
}

int main()
{
init();
while(scanf("%d", &n) != EOF) {
for(int i = 0; i <= n; ++i) {
scanf("%d", C + i);
}
scanf("%d", &m);
S = 0;
for(int i = 0, a; i < m; ++i) {
scanf("%d", &a);
S -= a;
S += p;
S %= p;
}
slove();
}
return 0;
}