我们记(deg(A))为多项式(A(x))的度，即为(A(x))的最高项系数 + 1
对于多项式(A(x))，如果存在(B(x))满足(deg(B) le deg(A))，且
[A(x)B(x) equiv 1 pmod {x^{n}}
]
我们称(B(x))为(A(x))在模(x^n)意义下的逆元，记作(A^{-1}(x))
求解过程##
考虑递归求解
当(n = 1)时，(A(x) equiv c pmod x)，显然(A^{-1}(x))就是(c^{-1})
倘若我们要计算
[A(x)B(x) equiv 1 pmod {x^n}
]
而已经计算出
[A(x)B'(x) equiv 1 pmod {x^{lceil frac{n}{2}
ceil}}
]
我们要求的(B(x))当然也满足
[A(x)B(x) equiv 1 pmod {x^{lceil frac{n}{2}
ceil}}
]
两式相减
[A(x)(B(x) - B'(x)) equiv 0 pmod {x^{lceil frac{n}{2}
ceil}}
]
即
[B(x) - B'(x) equiv 0 pmod {x^{lceil frac{n}{2}
ceil}}
]
两边平方，由于对于平方后的多项式(C(x))，其系数(c_i = sumlimits_{j = 0}^{i} b_j*b'_{i - j})，必有一项小于(lceil frac{n}{2}
ceil)而使(c_i = 0)
所以平方后放到(mod x^{n})意义下依然成立
[B^2(x) + B'^2(x) - 2B(x)B'(x) equiv 0 pmod {x^{n}}
]
两边乘(A(x))
[B(x) + A(x)B'^2(x) - 2B'(x) equiv 0 pmod {x^{n}}
]
得到
[B(x) equiv B'(x)(2 - A(x)B'(x)) pmod {x^{n}}
]
可以使用(fft)优化成(O(nlogn))
总时间复杂度(T(n) = T(lceil frac{n}{2}
ceil) + O(nlogn) = O(nlogn))
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 400005,maxm = 100005,INF = 1000000000;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
return out * flag;
}
const int G = 3,P = 998244353;
inline int qpow(int a,int b){
int re = 1;
for (; b; b >>= 1,a = 1ll * a * a % P)
if (b & 1) re = 1ll * re * a % P;
return re;
}
int a[maxn],b[maxn],c[maxn],R[maxn];
void NTT(int* a,int n,int f){
for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
for (int i = 1; i < n; i <<= 1){
int gn = qpow(G,(P - 1) / (i << 1));
for (int j = 0; j < n; j += (i << 1)){
int g = 1,x,y;
for (int k = 0; k < i; k++,g = 1ll * g * gn % P){
x = a[j + k]; y = 1ll * g * a[j + k + i] % P;
a[j + k] = (x + y) % P; a[j + k + i] = ((x - y) % P + P) % P;
}
}
}
if (f == 1) return;
int nv = qpow(n,P - 2); reverse(a + 1,a + n);
for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
void inv(int deg,int* a,int* b){
if (deg == 1){b[0] = qpow(a[0],P - 2); return;}
inv((deg + 1) >> 1,a,b);
int L = 0,n = 1;
while (n < (deg << 1)) n <<= 1,L++;
for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
for (int i = 0; i < deg; i++) c[i] = a[i];
for (int i = deg; i < n; i++) c[i] = 0;
NTT(c,n,1); NTT(b,n,1);
for (int i = 0; i < n; i++)
b[i] = 1ll * ((2ll - 1ll * c[i] * b[i] % P) + P) % P * b[i] % P;
NTT(b,n,-1);
for (int i = deg; i < n; i++) b[i] = 0;
}
int main(){
int n = read();
for (int i = 0; i < n; i++) a[i] = read();
inv(n,a,b);
for (int i = 0; i < n; i++) printf("%d ",b[i]);
return 0;
}