1 条题解

  • 0
    @ 2025-3-27 14:55:48
    #include <bits/stdc++.h>
    #define int long long
    using namespace std;
    int gcd(int a, int b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
    int t;
    int n, a[10000 + 5];
    signed main()
    {
        cin >> t;
        while (t--)
        {
            cin >> n;
            for (int i = 1; i <= n; i++)
                cin >> a[i];
            int now = a[2]; // 记录分母
            now /= gcd(a[1], now);
            for (int i = 3; i <= n; i++)
                now /= gcd(a[i], now);
            if (now == 1)
                cout << "Yes\n";
            else
                cout << "No\n";
        }
        return 0;
    }
    
    • 1

    信息

    ID
    3474
    时间
    1000ms
    内存
    128MiB
    难度
    2
    标签
    递交数
    17
    已通过
    5
    上传者