Codeforces Round #641 (Div. 2)

题目传送门

A-C(暂时) 在补了在补了

A. Orac and Factors

函数 f（x） 为x约数中大于1的最小的数，每一次操作使 n=n+f(n)，求k此操作后的n。

不难理解：f（n+f（n））=2，因为偶数约数一定有2，奇数的约数一定为奇数，

然后对于n为奇数，找一次f（n），剩下操作的一直加2。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
ll n, k;
 
int vis[1000010];
ll prime[1000010], id = 0;
 
void solve()
{
    cin >> n >> k;
    if (n & 1)
    {
        rep(i, 1, n) if (n % prime[i] == 0)
        {
            cout << n + prime[i] + 2 * (k - 1) << endl;
            break;
        }
    }
    else
        cout << n + k * 2 << endl;
}
int main()
{
 
    rep(i, 2, 1000010)
    {
        if (!vis[i])
            prime[++id] = i;
        for (int j = 1; j <= id && i * prime[j] <= 1000000; j++)
            vis[i * prime[j]] = 1;
    }
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

B. Orac and Models

求上升子序列，且满足相邻两项在原序列中的下标能被整除。

跟传统的上升子序列差不多。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
ll n, k;
ll s[100010], dp[100010], ans;
 
void solve()
{
    cin >> n;
    ans = 1;
    rep(i, 1, n)
    {
        cin >> s[i];
        dp[i] = 1;
    }
    rep(i, 1, n)
    {
        ans = max(ans, dp[i]);
        for (int j = i * 2; j <= n; j += i)
            if (s[i] < s[j])
                dp[j] = max(dp[j], dp[i] + 1);
    }
    cout << ans << endl;
}
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        solve();
    }
}

C. Orac and LCM

题意是s数组，求gcd（{lcm（{a_i，a_j}，i<j）}）

将a_i分解质因子p₁^k_i1,p₂^k_i2,,,

lcm（a_i,a_j）即为 p₁^{max(k_i1,k_j1)} * p₂^{max(k_i2,k_j2)} *...

gcd（a_i,a_j）即为 p₁^{min(k_i1,k_j1)} * p₂^{min(k_i2,k_j2)} *...

结合起来就是求p_x^{min（max(k_ix，k_jx)）}

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, a, b) for (register int i = a; i <= b; i++)
 
ll ksm(ll a, ll b)
{
    ll res = 1;
    for (; b; b >>= 1, a *= a)
        if (b & 1)
            res *= a;
    return res;
}
 
ll n, a[100010];
ll ans = 1;
 
int vis[1000010];
ll prime[1000010], id = 0;
 
void solve()
{
    cin >> n;
    rep(i, 1, n) cin >> a[i];
    for (int j = 1; j <= id; j++)
    {
        int m1 = 1000000, m2 = 1000000;
        for (int i = 1; i <= n; i++)
        {
            int tmp = 0;
            while (a[i] % prime[j] == 0)
            {
                tmp++;
                a[i] /= prime[j];
            }
            if (tmp < m1)
                swap(tmp, m1);
            if (tmp < m2)
                swap(tmp, m2);
            if (m2 == 0)
                break;
        }
        ans *= ksm(prime[j], m2);
    }
    cout << ans << endl;
}
int main()
{
    rep(i, 2, 200010)
    {
        if (!vis[i])
            prime[++id] = i;
        for (int j = 1; j <= id && i * prime[j] <= 200010; j++)
            vis[i * prime[j]] = 1;
    }
    int t;
    t = 1;
    while (t--)
    {
        solve();
    }
}