(FFT)

题意:

给一个函数\(f(x)=\sum_{i=0}^{n}c_i \cdot x^{i}\)

求\(g(x) = f(x - \sum a_i)\)后每一项\(x^{i}\)的系数mod998244353

\(n <= 10^{5},m <= 10^{5}\)

\(0 <= c_i < 998244353\)

\(0 <= a_i < 998244353\)

思路:

令\(d = -\sum a_i\),把\(g(x)\)展开得:

\[g(x) = c_0 (x + d)^{0} + c_1 (x + d)^{1} + ... + c_n (x+d)^{n}\]

令\(a_i = d^{i}\),再用二项式定理化简一下可以得到

\[g(x) = \sum_{i = 0}^{n}x^{i}\sum_{k=i}^{n}C(k,i)c_ka_{k-i}\]

\(fft\)只是入了门,想了半天,看不出来这是个卷积式子，组合数会变化啊,赛后终于开窍组合数是个阶乘啊,把\(c_k 和 a^{k-i}变换一下\)

令\[b_k = c_k \cdot k!, a_i = \frac{a_i}{i!}\]

\(g(x)\)就可以写成

\[g(x) = \sum_{i = 0}^{n}x^{i} \cdot \frac{1}{i!}\sum_{k=i}^{n}b_ka_{k-i}\]

令\(ans(i) = \frac{1}{i!} \sum_{k=i}^{n}b_ka_{k-i}\)

把b数组逆序一下

\(ans(i) = \frac{1}{i!}\sum_{k=0}^{n-i}b_{k}a_{n-k-i}\)

类比fft多项式乘法下面\(c_j\)的形式 \(\sum_{k=0}^{n-i}b_{k}a_{n-k-i}\)这一项其实就是\(fft\)之后得到的数组\(c_{n-i}\),最后答案\(ans(i) = \frac{1}{i!} c_{n-i}\)

\(A(x) = \sum_{i=0}^{n}a_ix^{i}\)

\(B(x) = \sum_{i=0}^{n}b_ix^{i}\)

\(C(x) = A(x)B(x) = \sum_{i=0}^{2n}c_ix^{i}\)

\(c_j = \sum_{i=0}^{j}a_ib_{j-i}\)

然后就上板子了，由于是在模意义下的运算,要拿ntt,去找了个板子

不太会用啊，板子上的费马素数是P=(1LL<<55) * 5+1,原根g=6的，

开始交了几发，TLE，原来是数组开小了，改完再交RE了，也不知道改了哪里就没RE了，然后WA了，暴力对拍数据，发现是费马素数的锅，乱试了其他的一些费马素数,又想了半天觉得这样不行，本来就是在mod下取的逆元，又在P下做运算，ntt原理也不懂，一脸懵逼,最后我直接把P改成mod试了一下，居然A了，好像给的这个mod本来是就是一个费马素数(1<<23) * 119 + 1,g = 3，而且运气好前面试的费马素数原根刚好是3。

还有疑问就是运算时费马素数应该取多大呢,==再深入学习一下

#include
    
     #define LL long longusing namespace std;const int N = 2e5 + 1000;const int mod = 998244353;const LL P =  mod;const LL G = 3;const int NUM = 23;int read(){    int x = 0;    char c;    while((c=getchar())<'0'||c>'9');    while(c>='0'&&c<='9')        x=x*10+(c-'0'), c=getchar();    return x;}int fac[N],facinv[N];int n, m;LL mul(LL x,LL y){ //return (x * y - (LL)(x / (long double)P * y + 1e-3) * P + P) % P; return x * y % P;}LL q_pow(LL a,LL b){  LL res = 1,tmp = a;  while(b){    if(b &1) res = res * tmp % P;    tmp = tmp * tmp % P;    b >>= 1;  }  return res;}void init(){    fac[0] = facinv[0] = 1;    for(int i = 1;i < N;i++){        fac[i] = 1LL * i * fac[i-1] % mod;        facinv[i] = 1LL * q_pow(i, mod - 2) * facinv[i - 1] % mod;    }}LL  wn[NUM];LL  a[2 * N], b[2 * N],c[N];void GetWn(){    for(int i = 0; i< NUM; i++)    {        int t = 1 << i;        wn[i] = q_pow(G, (P - 1) / t);    }}void Rader(LL a[], int len){    int j = len >> 1;    for(int i=1; i
     
      > 1;        while(j >= k)        {            j -= k;            k >>= 1;        }        if(j < k) j += k;    }}void NTT(LL a[], int len, int on){    Rader(a, len);    int id = 0;    for(int h = 2; h <= len; h <<= 1)    {        id++;        for(int j = 0; j < len; j += h)        {            LL w = 1;            for(int k = j; k < j + h / 2; k++)            {                LL u = a[k];                LL t = mul(w,a[k + h / 2]);                a[k] = (u + t) % P;                a[k + h / 2] = ((u - t) % P + P) % P;                w = mul(w,wn[id]);            }        }    }    if(on == -1)    {        for(int i = 1; i < len / 2; i++)            swap(a[i], a[len - i]);        LL Inv = q_pow(len, P - 2);        for(int i = 0; i < len; i++)            a[i] = mul(a[i],Inv);    }}void Conv(LL a[], LL b[], int n){    NTT(a, n, 1);    NTT(b, n, 1);    for(int i = 0; i < n; i++) a[i] = mul(a[i],b[i]);    NTT(a, n, -1);}int main(){    GetWn();    init();    while(scanf("%d",&n) == 1){        for(int i = 0;i <= n;i++) c[i] = read();        int  sum = 0;        m = read();        for(int i = 1;i <= m;i++){            int x;            x = read();            sum = (sum - x + mod) % mod;        }        int len = 1;        while(len < 2 * (n + 1)) len <<= 1;        int res = 1;        for(int i = 0;i <= n;i++) {            a[i] = 1LL * res * facinv[i] % mod, res = 1LL * res * sum % mod;            b[i] = c[n - i] * fac[n - i] % mod;        }        for(int i = n + 1;i < len;i++) a[i] = b[i] = 0;        Conv(a,b,len);        for(int i = 0;i <= n;i++) printf("%lld ",a[n - i] * facinv[i] % mod);        printf("\n");    }    return 0;}