题解：
- (F[n+m]=F[n]*F[m]+F[n-1]*F[m-1])
- (F[n]*F[m]=F[n+m]-(F[n-1]*F[m-1]))
- (...)
- (F[n]*F[m]=F[n+m]-F[n+m-2]+F[n+m-4]…+(-1)^{min(n,m)}*F[|n-m|])
用FFT求(A(x)*B(x))，再求(A(-x)*B(frac{1}{x}))，即可得到答案是两个斐波拉契表示（非01最简表示）的差。
考虑把一个斐波拉契表示（非01最简）化简成01表示，即化简(sum_{i=1}^n a[i]*F[i])，可以用分治：
(sum_{i=1}^n a[i]*F[i]=sum_{i=1}^n a[i]~mod~2*F[i]+2(sum_{i=1}^n lfloor frac{a[i]}{2} floor*F[i]))
问题在于两个01最简表示做怎么(O(n))做加法？
- 考虑倒着做，每次在(i)这个位置加(a[i]+b[i])次1，不难发现复杂度挺对的。
那么最后的问题：做差
- 还是倒着做，如果(i)位置要(-1)，找到最小的(j)满足(a[j]=1)且(jge i)，发现是把它-1，然后给(i-1..j-2~)加1，不难发现复杂度还是对的。
时间复杂度：(O(n~log~n))
Code：
#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, _b = y; i <= _b; i ++)
#define ff(i, x, y) for(int i = x, _b = y; i < _b; i ++)
#define fd(i, x, y) for(int i = x, _b = y; i >= _b; i --)
#define ll long long
#define pp printf
#define hh pp("
")
using namespace std;
const int nm = 1 << 21;
#define db double
namespace fft {
const db pi = acos(-1);
struct P {
db x, y;
P(db _x = 0, db _y = 0) {
x = _x, y = _y;
}
};
P operator + (P a, P b) { return P(a.x + b.x, a.y + b.y);}
P operator - (P a, P b) { return P(a.x - b.x, a.y - b.y);}
P operator * (P a, P b) { return P(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x);}
P w[nm]; int r[nm];
void build() {
for(int i = 1; i < nm; i *= 2) {
ff(j, 0, i) {
w[i + j] = P(cos(pi * j / i), sin(pi * j / i));
}
}
}
void dft(P *a, int n, int f) {
ff(i, 0, n) {
r[i] = r[i / 2] / 2 + (i & 1) * (n / 2);
if(i < r[i]) swap(a[i], a[r[i]]);
} P b;
for(int i = 1; i < n; i *= 2) for(int j = 0; j < n; j += 2 * i) ff(k, 0, i)
b = a[i + j + k] * w[i + k], a[i + j + k] = a[j + k] - b, a[j + k] = a[j + k] + b;
if(f == -1) {
reverse(a + 1, a + n);
ff(i, 0, n) a[i].x /= n, a[i].y /= n;
}
}
P c[nm], d[nm], p[nm];
P conj(P a) { return P(a.x, -a.y);}
void mtp(ll *a, ll *b, int n) {
ff(i, 0, n) p[i] = P(a[i], b[i]);
dft(p, n, 1);
ff(i, 0, n) {
P k = conj(p[(n - i) % n]);
c[i] = (p[i] + k) * P(0.5, 0);
d[i] = (p[i] - k) * P(0, -0.5);
c[i] = c[i] * d[i];
}
dft(c, n, -1);
ff(i, 0, n) a[i] = round(c[i].x);
}
}
using fft :: mtp;
int n, m;
ll a[nm], b[nm], a0[nm], b0[nm], c0[nm];
ll c[nm], d[nm];
ll p[nm], q[nm];
void add(ll *f, int x) {
if(x == 0) {
add(f, x + 1);
return;
}
if(f[x + 1]) {
f[x + 1] = 0;
add(f, x + 2);
return;
}
if(f[x - 1]) {
f[x - 1] = 0;
add(f, x + 1);
return;
}
if(f[x]) {
f[x] = 0;
if(x == 1) {
add(f, x + 1);
return;
}
add(f, x - 2); add(f, x + 1);
return;
}
f[x] = 1;
}
void plu(ll *p, ll *q, int n) {
static ll f[nm];
fo(i, 0, n) f[i] = 0;
fd(i, n, 1) {
fo(j, 1, p[i] + q[i]) add(f, i);
}
fo(i, 0, n) p[i] = f[i];
}
bool bz[35][nm];
void solve(int D, ll *c, ll *p, int n) {
ll mx = 0;
fo(i, 1, n) mx = max(mx, c[i]);
if(mx == 0) {
fo(i, 1, n) p[i] = 0;
return;
}
fo(i, 1, n) {
bz[D][i] = c[i] & 1;
c[i] /= 2;
}
solve(D + 1, c, p, n);
plu(p, p, n);
static ll g[nm];
fo(i, 1, n) g[i] = bz[D][i];
plu(p, g, n);
}
void dec(ll *p, ll *q, int n) {
fd(i, n, 1) if(q[i]) {
int l = i;
while(!p[l]) l ++;
p[l] --;
fd(j, l - 2, i - 1) add(p, j);
}
}
void work() {
scanf("%d", &n);
fo(i, 1, n) scanf("%lld", &a[i]);
scanf("%d", &m);
fo(i, 1, m) scanf("%lld", &b[i]);
int tp = 1;
while(1 << ++ tp <= n + m);
ff(i, n + 1, 1 << tp) a[i] = 0;
ff(i, m + 1, 1 << tp) b[i] = 0;
ff(i, 0, 1 << tp) a0[i] = a[i], b0[i] = b[i];
mtp(a, b, 1 << tp);
// fo(i, 1, n) if(a0[i]) fo(j, 1, m) if(b0[j]) {
// int w = abs(i - j);
// c0[w] += min(i, j) % 2 ? -1 : 1;
// }
fo(i, 1, n) a0[i] = a0[i] * (i % 2 ? -1 : 1);
reverse(b0 + 0, b0 + m + 1);
mtp(a0, b0, 1 << tp);
ff(i, 0, (1 << tp) + 100) c0[i] = 0;
ff(i, 0, 1 << tp) if(a0[i]) {
int j = abs(i - m);
if(i - m <= 0) {
c0[j] += a0[i];
} else {
c0[j] += a0[i] * (j % 2 ? -1 : 1);
}
}
n = (1 << tp) + 100;
fo(i, 0, (1 << tp) + 100) c[i] = d[i] = 0;
fd(i, (1 << tp) - 1, 0) {
c[i] = a[i] + d[i + 2];
d[i] = c[i + 2];
if(c0[i + 2] > 0) c[i] += c0[i + 2]; else
d[i] -= c0[i + 2];
}
c[1] += c[0]; c[0] = 0;
d[1] += d[0]; d[0] = 0;
fo(i, 0, n) p[i] = q[i] = 0;
solve(0, c, p, n);
solve(0, d, q, n);
m = 0;
dec(p, q, n);
fo(i, 1, n) if(p[i]) m = i;
pp("%d ", m);
fo(i, 1, m) pp("%lld ", p[i]); hh;
}
int main() {
fft :: build();
int T; scanf("%d", &T);
fo(ii, 1, T) {
work();
}
}