E. G-C-D, Unlucky!

https://codeforces.com/problemset/problem/2126/E

题意：给定一个未知数组a的前缀和后缀gcd数组p和s，问这样的数组a是否存在，输出yes或者no。

思路：考虑a中的每一个位置i，gcd(a[i], s[i - 1]) == g1, gcd(a[i], p[i + 1]) == g2, 那么a[i]必定包含lcm(g1, g2)，这样才能满足条件，根据此逻辑，求出所有的a，然后做一次运算，看能否得到p和s即可。

inline void solve() {
int n;
cin >> n;

vector p(n), s(n);
for (auto& x : p) {
cin >> x;
}

for (auto& x : s) {
cin >> x;
}

if (p[n - 1] != s[0]) {
cout <<"NO\n";
return;
}

auto valid = [](const std::vector& a) {
int n = (int)a.size();
for (int i = 0; i < n - 1; ++i) {
if (a[i] % a[i + 1] != 0) {
return false;
}
}
return true;
};

reverse(s.begin(), s.end());
if (!valid(p) || !valid(s)) {
cout <<"NO\n";
return;
}

reverse(s.begin(), s.end());
vector a(n);

a[0] = p[0];
a[n - 1] = s[n - 1];
for (int i = 1; i < n - 1; ++i) {
a[i] = ::lcm(p[i], s[i]);
}

long long g = 0;
for (int i = 0; i < n; ++i) {
g = ::gcd(g, a[i]);
if (g != p[i]) {
cout << "No\n";
return;
}
}

g = 0;
for (int i = n - 1;i >= 0; --i) {
g = ::gcd(g, a[i]);
if (g != s[i]) {
cout << "NO\n";
return;
}
}

cout << "Yes\n";
}