[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

題解

待求表達式可以拆開然後化成 a(n)=∑ni=0h(i)g(i) 的形式

注意到隻要令 h′(x)=h(n−x) 就可到一個卷積式 (h⊗g)(n)=∑ni=0h′(n−i)g(i)

然後NTT

#include<stdio.h>

typedef long long LL;
const int MAXN =  << ;
const int p    = ( << ) + ;
const int g    = ;
const int NUM  = ;
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 = ;
    while(k) {
        if(k&) res = L * res * a % p;
        a = L * a * a % p;
        k >>= ;
    }
    return res;
}

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

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

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

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

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