挺好的一道题目。像这种分久必合,合久必分的题目,我们可以考虑把能分解的都分解。把 $a$ 中的每个元素都分解到不能分解,对 $b$ 中的每个元素,考虑是否可以由分解后 $a$ 的前缀元素组成。有一个更加好写的方法:把 $b$ 中的每个元素也分解到不能分解,然后比较两个分解后的数组是否相等。
有一个陷阱:如果对分解后的 $a$ 数组合并同类项,可能会出现整数溢出,需要使用 long long
类型。
// Date: Sat Feb 10 17:00:44 2024
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;
template <class T> using pq = priority_queue<T>;
template <class T> using pqg = priority_queue<T, vector<T>, greater<T>>;
const int INF = 0x3f3f3f3f, MOD = 1e9 + 7, MOD1 = 998'244'353;
const ll INFL = 0x3f3f3f3f'3f3f3f3f;
const double eps = 1e-8;
const int dir[8][2] = {
{0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};
const ull Pr = 131;
#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])
#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nemp(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())
#define NL cout << '\n';
template <class T> bool ckmin(T &a, const T &b) { return b < a ? a = b, 1 : 0; }
template <class T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
template <typename t> istream &operator>>(istream &in, vector<t> &vec) {
for (t &x : vec)
in >> x;
return in;
}
template <typename t> ostream &operator<<(ostream &out, vector<t> &vec) {
int n = SZ(vec);
For(i, 0, n) {
out << vec[i];
if (i < n - 1)
out << ' ';
}
return out;
}
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V> void __print(const pair<T, V> &x) {
cerr << '{';
__print(x.first);
cerr << ", ";
__print(x.second);
cerr << '}';
}
template <typename T> void __print(const T &x) {
int f = 0;
cerr << '{';
for (auto &i : x)
cerr << (f++ ? ", " : ""), __print(i);
cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V> void _print(T t, V... v) {
__print(t);
if (sizeof...(v))
cerr << ", ";
_print(v...);
}
#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z) \
cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#define dbg(x...) \
cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \
_print(x); \
cerr << "\e[39m" << endl;
#else
#define debug1
#define debug2
#define debug3
#define dbg(x...)
#endif
const int N = 50100;
int n, m, k, a[N], b[N];
PII split(int x) {
int cnt = 1;
while (x % m == 0) {
cnt *= m;
x /= m;
}
return {x, cnt};
}
using PLL = pair<ll, ll>;
void solve() {
cin >> n >> m;
For1(i, 1, n) { cin >> a[i]; }
cin >> k;
For1(i, 1, k) { cin >> b[i]; }
vector<PII> d, d1;
For1(i, 1, n) { d.pb(split(a[i])); }
For1(i, 1, k) { d1.pb(split(b[i])); }
auto shrink = [](vector<PII> &d) {
vector<PLL> da;
ll pre = -1, cnt = 0;
for (auto &[x, c] : d) {
if (x == pre) {
cnt += c;
} else if (pre == -1) {
pre = x;
cnt = c;
} else {
da.pb({pre, cnt});
pre = x, cnt = c;
}
}
da.pb({pre, cnt});
return da;
};
vector<PLL> da = shrink(d), db = shrink(d1);
bool flag = true;
if (SZ(da) != SZ(db))
flag = false;
else {
int len = SZ(da);
For(i, 0, len) {
if (da[i] != db[i]) {
flag = false;
break;
}
}
}
cout << (flag ? "YES" : "NO") << '\n';
}
int main(void) {
#ifdef _DEBUG
freopen("1696c.in", "r", stdin);
#endif
std::ios::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int T = 1;
cin >> T;
while (T--) {
solve();
}
return 0;
}