Codefores 1025D Recovering BST

给定一个序列，问这个序列是否能够构成一个二叉搜索树，使得任一边连接的点的gcd大于1

区间dp

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;

const int maxn = 710;

int A[maxn];
bool G[maxn][maxn];
bool dp[maxn][maxn][2];

int gcd(int x, int y) {
	return y == 0 ? x : gcd(y, x % y);
}

int main() {
	int n; scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
	for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) {
		G[i][j] = (gcd(A[i], A[j]) > 1);
	}
	for (int len = 1; len <= n; len++) {
		for (int l = 1, r = l + len - 1; l + len - 1 <= n; l++, r++) {
			for (int m = l; m <= r; m++) {
				bool flag = true;
				if (l == m) {
					flag &= true;
				} else {
					flag &= dp[l][m - 1][0];
				}
				if (m == r) {
					flag &= true;
				} else {
					flag &= dp[m + 1][r][1];
				}
				if (flag) {
					dp[l][r][1] |= G[m][l - 1];
					dp[l][r][0] |= G[m][r + 1];
				}
			}
		}
	}

	for (int i = 1; i <= n; i++) {
		bool tag = true;
		if (i == 1) {
			tag &= true;
		} else {
			tag &= dp[1][i - 1][0];
		}
		if (i == n) {
			tag &= true;
		} else {
			tag &= dp[i + 1][n][1];
		}
		if (tag) {
			puts("Yes"); return 0;
		}
	}
	puts("No");
	return 0;
}