Good Bye 2023

A. 2023

思路：如果这些数是2023的因子，只需用1补全原序列即可

code:

void solved()
{
    int a = 2023, flag = 0;
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; ++i)
    {
        int temp;
        cin >> temp;
        if (a % temp != 0 && flag == 0)
        {
            cout << "NO" << endl;
            flag = 1;
        }
        else
        {
            a /= temp;
        }
    }
    if (flag == 0)
    {
        cout << "YES" << endl;
        cout << a << " ";
        for (int i = 1; i < k; ++i)
            cout << 1 << " ";
        cout << endl;
    }
}

B. Two Divisors

思路：找出a和b中最小的质因子，和b相乘即可

code:

void Eratosthenes(int n)
{
    for (LL i = 2; i <= n; ++i)
    {
        if (!vis[i])
        {
            prim[cnt++] = i;
            for (LL j = i * i; j <= n; j += i)
                vis[j] = 1;
        }
    }
}

void solved()
{
    int a, b;
    cin >> a >> b;
    for (int i = 0; i < cnt; ++i)
    {
        if (a % prim[i] == 0 || b % prim[i] == 0)
        {
            cout << b * prim[i] << endl;
            return;
        }
    }
}

C. Training Before the Olympiad

思路：两个人无论谁操作都会将奇数变偶数，若有奇数，Olya会使总数减一，Masha可以在一次操作中消去两个奇数，只有一个奇数时，无论谁操作都会使总数减一，要开long long

code:

void solved()
{
    cin >> n;
    for (int i = 1; i <= n; ++i)
    {
        cin >> a[i];
    }
    int odd = 0, even = 0, sum = 0;
    for (int i = 1; i <= n; ++i)
    {
        sum += a[i];
        if (a[i] & 1)
            odd++;
        if (i == 1)
        {
            cout << sum << " ";
        }
        else
        {
            if (odd > 0)
            {
                int ans = sum - odd / 3;
                if (odd % 3 == 1)
                    ans--;
                cout << ans << " ";
            }
            else
            {
                cout << sum << " ";
            }
        }
    }
    cout << endl;
}

D. Mathematical Problem

思路：可以通过打表发现规律，1,6,9可以构造

code:

void solved()
{
    int n;
    cin >> n;
    if (n == 1)
    {
        cout << 1 << endl;
        return;
    }
    cout << 196 << string(n - 3, '0') << endl;
    for (int i = 0; i < n / 2; ++i)
    {
        cout << 9 << string(i, '0') << 6 << string(i, '0') << 1 << string(n - 3 - 2 * i, '0') << endl;
        cout << 1 << string(i, '0') << 6 << string(i, '0') << 9 << string(n - 3 - 2 * i, '0') << endl;
    }
}