HDU 6061 RXD and functions

题目链接：HDU-6061

题意：给定f(x)，求f(x-A)各项系数。

思路：推导公式有如下结论：

然后用NTT解决即可。

代码：

#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
typedef long long LL;
const LL MAXN=;
const LL MOD=;

// NTT
// O(nlogn)
// Verified!
const LL P = MOD;
const LL G = ;
const LL NUM = ;

LL  wn[NUM];
LL A[MAXN], B[MAXN];

LL quick_mod(LL a, LL b, LL m)
{
    LL ans = ;
    a %= m;
    while(b)
    {
        if(b & )
        {
            ans = ans * a % m;
            b--;
        }
        b >>= ;
        a = a * a % m;
    }
    return ans;
}

void GetWn()
{
    for(LL i=; i<NUM; i++)
    {
        LL t =  << i;
        wn[i] = quick_mod(G, (P - ) / t, P);
    }
}

void Prepare(char A[], char B[], LL a[], LL b[], LL &len)
{
    len = ;
    LL len_A = strlen(A);
    LL len_B = strlen(B);
    while(len <=  * len_A || len <=  * len_B) len <<= ;
    for(LL i=; i<len_A; i++)
        A[len -  - i] = A[len_A -  - i];
    for(LL i=; i<len - len_A; i++)
        A[i] = '';
    for(LL i=; i<len_B; i++)
        B[len -  - i] = B[len_B -  - i];
    for(LL i=; i<len - len_B; i++)
        B[i] = '';
    for(LL i=; i<len; i++)
        a[len -  - i] = A[i] - '';
    for(LL i=; i<len; i++)
        b[len -  - i] = B[i] - '';
}

void Rader(LL a[], LL len)
{
    LL j = len >> ;
    for(LL i=; i<len-; i++)
    {
        if(i < j) swap(a[i], a[j]);
        LL k = len >> ;
        while(j >= k)
        {
            j -= k;
            k >>= ;
        }
        if(j < k) j += k;
    }
}

void NTT(LL a[], LL len, LL on)
{
    Rader(a, len);
    LL id = ;
    for(LL h = ; h <= len; h <<= )
    {
        id++;
        for(LL j = ; j < len; j += h)
        {
            LL w = ;
            for(LL k = j; k < j + h / ; k++)
            {
                LL u = a[k] % P;
                LL t = w * (a[k + h / ] % P) % P;
                a[k] = (u + t) % P;
                a[k + h / ] = ((u - t) % P + P) % P;
                w = w * wn[id] % P;
            }
        }
    }
    if(on == -)
    {
        for(LL i = ; i < len / ; i++)
            swap(a[i], a[len - i]);
        LL Inv = quick_mod(len, P - , P);
        for(LL i = ; i < len; i++)
            a[i] = a[i] % P * Inv % P;
    }
}

void Conv(LL a[], LL b[], LL n)
{
    NTT(a, n, );
    NTT(b, n, );
    for(LL i = ; i < n; i++)
        a[i] = a[i] * b[i] % P;
    NTT(a, n, -);
}

// 快速幂
// 求x^n%mod
// Verified!
LL powMod(LL x,LL n,LL mod)
{
    LL res=;
    while(n>)
    {
        if(n&) res=res*x % mod;
        x=x*x % mod;
        n>>=;
    }
    return res;
}
// 求逆元
// a和ｍ应该互质
// 根据欧拉定理：a的逆即a^(phi(m)-1)
LL inv(LL a,LL m)
{
    return powMod(a,m-,m);
    // return powMod(a,eularPhi(m)-1,m);
}
LL mi[MAXN],invsum[MAXN],fac[MAXN];
LL c[MAXN];
int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    GetWn();
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        n++;
        for(int i=;i<n;i++) scanf("%lld",&c[i]);
        int m,s=;
        scanf("%d",&m);
        for(int i=;i<=m;i++)
        {
            int tmp;
            scanf("%d",&tmp);
            s=(s+tmp)%MOD;
        }
        int len=;
        while(len<*n) len*=;
        mi[]=fac[]=invsum[]=invsum[]=;
        for(int i=;i<n;i++)
            mi[i]=mi[i-]*(P-s)%MOD;
        for(int i=;i<n;i++)
            fac[i]=fac[i-]*i%MOD;
        for(int i=;i<n;i++)
            invsum[i]=inv(fac[i],MOD);
        for(int i=;i<n;i++)
            A[i]=mi[i]*invsum[i]%MOD;
        for(int i=;i<n;i++)
            B[i]=c[n-i-]*fac[n-i-]%MOD;
        for(int i=n;i<len;i++)
            A[i]=B[i]=;
        Conv(A, B, len);
        for(int i=;i<n;i++)
            printf("%lld ",A[n-i-]*invsum[i]%MOD);
        printf("\n");
    }
    return ;
}