HDU 6061 RXD and functions NTT

RXD and functions

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

$(x - 1) ^ 2 = x^2 - 2x + 1$

 

题解：

　　

代码：

　　

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
typedef unsigned long long ULL;
const long long INF = 1e18+1LL;
const double pi = acos(-1.0);
const int N = 5e5+, M = 1e3+,inf = 2e9;

const long long P=998244353LL,mod = 998244353LL;
const LL G=3LL;

LL mul(LL x,LL y){
    return (x*y-(LL)(x/(long double)P*y+1e-)*P+P)%P;
}
LL qpow(LL x,LL k){
    LL ret=;
    while(k){
        if(k&) ret=mul(ret,x);
        k>>=;
        x=mul(x,x);
    }
    return ret;
}
LL wn[];
void getwn(){
    for(int i=; i<=; ++i){
        int t=<<i;
        wn[i]=qpow(G,(P-)/t);
    }
}

int len;
void NTT(LL y[],int op){
    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;
    }
    int id=;
    for(int h=; h<=len; h<<=) {
        ++id;
        for(int i=; i<len; i+=h){
            LL w=;
            for(int j=i; j<i+(h>>); ++j){
                LL u=y[j],t=mul(y[j+h/],w);
                y[j]=u+t;
                if(y[j]>=P) y[j]-=P;
                y[j+h/]=u-t+P;
                if(y[j+h/]>=P) y[j+h/]-=P;
                w=mul(w,wn[id]);
            }
        }
    }
    if(op==-){
        for(int i=; i<len/; ++i) swap(y[i],y[len-i]);
        LL inv=qpow(len,P-);
        for(int i=; i<len; ++i) y[i]=mul(y[i],inv);
    }
}
LL c[N],fac[N],ans[N],inv[N],id[N],s[N],t[N];
int n;
void solve(LL mo) {
    if(mo == ) {
        for(int i = ; i <= n; ++i)
            ans[i] = c[i];
        return ;
    }
    mo = (mod - mo) % mod;
    len = ;
    while(len <= *n+) len<<=;
    id[] = ;
    for(int i = ; i <= n; ++i)
        id[i] = id[i-] * mo % mod;
    for(int i = ; i < len ; ++i) s[i] = ,t[i] = ;
    for(int i = ; i <= n; ++i)
        s[i] = c[i]*fac[i]%mod,
        t[n - i] = id[i] * inv[i] % mod;
    NTT(s,),NTT(t,);
    for(int i = ; i < len; ++i) s[i] = s[i]*t[i] % mod;
    NTT(s,-);
    for(int i = ; i <= n; ++i) {
        ans[i] = s[n+i]*inv[i] % mod;
    }
}
int m;
int main() {
    getwn();
    while(scanf("%d",&n)!=EOF) {
        for(int i = ; i <= n; ++i) {
            scanf("%lld",&c[i]);
        }
        fac[] = ;
        for(int i = ; i <= n; ++i) {
            fac[i] = fac[i-]*1LL*i%mod;
        }
        inv[n]=qpow(fac[n],mod-);
        for(int i = n-; i >= ; --i)
            inv[i]=inv[i+]*1ll*(i+)%mod;
        scanf("%d",&m);
        int sum = ;
        for(int i = ; i <= m; ++i) {
            int x;
            scanf("%d",&x);
            sum += x;
            sum %= mod;
        }
        solve(sum);
        for(int i = ; i < n; ++i)
            printf("%lld ",ans[i]);
        printf("%lld \n",ans[n]);
    }
    return ;
}