**Preface:**
Recently wanted to write a template for polynomial modulo, but could not find the template problem. So the internet to find a high-precision division problem, after writing to find that the high-precision division and polynomial modulus is not the same qaq. The former requires no negative numbers, for which some positions can be temporarily 0, the remainder is pulled to the back, while polynomial modulus requires that the highest item must remain 0.

So I changed the program casually, first throw it here, after the memo. I'm writing a real number (FFT) version (I don't know what the real version of the polynomial modulo actually does).

In addition I do not know whether to write wrong, after a few groups of hand-made small data on the ...

CODE:

#include <iostream> #include <string> #include <cstring> #include <cmath> #include <cstdio
> #include <cstdlib> #include <stdio.h> #include <algorithm> using namespace std;
const int maxn=1000000;
Const double Pi=acos (-1.0);
struct Complex {double x, y;
Complex (Double a=0.0,double b=0.0): X (a), Y (b) {}};
Complex operator+ (Complex A,complex b) {return Complex (A.X+B.X,A.Y+B.Y);}
Complex operator-(Complex A,complex b) {return Complex (A.X-B.X,A.Y-B.Y);}
Complex operator* (Complex A,complex b) {return Complex (a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
Complex A[MAXN];
Complex B[MAXN];
int REV[MAXN];
int N,LG;
Double F[MAXN];
int G[MAXN];
Double ANS[MAXN];
Double MOD[MAXN];
Char S[MAXN];
int H[MAXN];
int ln,lm;
void DFT (Complex *a,double f) {for (int i=0; i<n; i++) if (I<rev[i]) swap (a[i],a[rev[i]);
for (int len=2; len<=n; len<<=1) {int mid= (len>>1);
Double ang=2.0*pi/(double) Len; Complex e (cos (ANG), F*sin (ANG));
for (Complex *p=a; p!=a+n; p+=len) {Complex wn (1.0,0.0);
for (int i=0; i<mid; i++) {Complex temp=wn*p[mid+i];
P[mid+i]=p[i]-temp;
P[i]=p[i]+temp;
Wn=wn*e;
}}}} void FFT (bool rev) {for (int i=0; i<n; i++) {rev[i]=0;
for (int j=0; j<lg; J + +) if (i& (1<<j)) rev[i]|= (1<< (lg-j-1));
} DFT (a,1.0);
DFT (b,1.0);
if (rev.) for (int i=0; i<n; i++) a[i]=a[i]*a[i];
for (int i=0; i<n; i++) a[i]=a[i]*b[i];
DFT (a,-1.0); for (int i=0; i<n; i++) a[i].
X/= (double) N;
} void Poly_rev (int m) {if (m==1) f[0]=1.0/(double) g[0];
else {Poly_rev (m>>1);
n=2*m,lg=0;
while ((1<<LG) <n) lg++;
for (int i=0; i<m; i++) A[i]=complex (f[i],0.0), B[i]=complex (g[i],0.0); FoR (int i=m; i<n; i++) A[i]=b[i]=complex (0.0,0.0);
FFT (TRUE); for (int i=0; i<m; i++) f[i]=2.0*f[i]-a[i].
X
}} int main () {freopen ("c.in", "R", stdin);
Freopen ("C.out", "w", stdout);
scanf ("%d", &LN);
for (int i=ln-1; i>=0; i--) scanf ("%d", &h[i]);
scanf ("%d", &LM);
for (int i=lm-1; i>=0; i--) scanf ("%d", &g[i]);
N=1;
while (n<ln-lm+3) n<<=1;
Poly_rev (N);
ln=ln-lm+1;
n=1,lg=0;
while (n<2*ln+2) n<<=1,lg++;
for (int i=0; i<ln; i++) A[i]=complex (f[i],0.0), B[i]=complex (h[i],0.0);
for (int i=ln; i<n; i++) A[i]=b[i]=complex (0.0,0.0);
FFT (FALSE); for (int i=0; i<ln; i++) ans[i]=a[i].
X
for (int i=0; i<ln-i-1; i++) swap (ans[i],ans[ln-i-1]);
for (int i=0; i<ln; i++) printf ("%.3lf", Ans[i]);
printf ("\ n");
n=1,lg=0;
while (n<ln+lm+2) n<<=1,lg++;
for (int i=0; i<n; i++) A[i]=b[i]=complex (0.0,0.0); for (int i=0; i<ln
i++) A[i]=complex (ans[i],0.0);
for (int i=0; i<lm; i++) B[i]=complex (g[lm-i-1],0.0);
FFT (FALSE); for (int i=0; i<lm-1; i++) mod[i]=h[ln+lm-i-2]-a[i].
X
for (int i=0; i<lm-1; i++) printf ("%.3lf", Mod[i]);
printf ("\ n");
return 0; }