B. Playing with GCD

传送门

题意:
一个长度为n的数组a, $a_i = gcd(b_i, b_{i + 1})$, 问是否存在这样的b数组能够构成a

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思路:

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总结:
gcd可以推导出lcm的规律，图片中的那个 >= 关系是代表要产生$a_i$，必须最优是要满足这个关系，但是不一定满足了这个关系就能达成目的

点击查看代码

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false);
#define endl '\n'
using namespace std;

typedef long long ll;
const int MAXN = 1e5 + 10;
int T, n;
int a[MAXN], b[MAXN];

int gcd(int a, int b)
{
    return b == 0 ? a : gcd(b, a % b);
}

int lcm(int a, int b)
{
    return  a * b / gcd(a, b);
}

int main()
{
    IOS; cin.tie(0), cout.tie(0);
    cin >> T;
    while (T--)
    {
        cin >> n;
        for (int i = 1; i <= n; ++i)
            cin >> a[i];
        b[1] = a[1], b[n + 1] = a[n];
        for (int i = 2; i <= n; ++i)
            b[i] = lcm(a[i - 1], a[i]);
        bool flag = true;
        for (int i = 1; i <= n; ++i)
            if (a[i] != gcd(b[i], b[i + 1]))
                flag = false;
        cout << (flag ? "YES" : "NO") << endl;        
    }
    return 0;
}