CF1285F Classical?

发表于 | 阅读数

数列$a_{1..n}$

$lcm(a,b)=\frac{a*b}{gcd(a,b)}$

枚举$gcd(a,b)<=\sqrt n$,构成$gcd$,$a_i/gcd=b_i$,变成了求$b_i$中乘积最大互质。

将b_i从大到小排序，设置一个栈$s$，遍历$b_i$,设当前遍历为$x$?

  • $ans=max(ans,xk(栈内与x互质的最大数)g)$
  • 贪心的发现比$k$小栈内的数都失去了机会。

如何找到$k$,可以通过找到栈内有多少与$k$互质的数。

一个数的因数个数$\log n$

复杂度$O(sqrt(n)nlog(n))$ ，实现：$600ms\leq 1000ms$

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
const int mod = 1e9 + 7;
vector<int> d[N];
int prime[N], notprime[N], mu[N], cnt;
int gcd(int a, int b)
{
    return b == 0 ? a : gcd(b, a % b);
}
void init()
{
    for (int i = 1; i < N; i++)
        for (int j = 1; j * j <= i; j++)
            if (i % j == 0)
            {
                d[i].push_back(j);
                if (j * j != i)
                    d[i].push_back(i / j);
            }
    mu[1] = 1;
    for (int i = 2; i < N; i++)
    {
        if (!notprime[i])
        {
            prime[++cnt] = i;
            mu[i] = -1;
        }
        for (int j = 1; j <= cnt && i * prime[j] < N; j++)
        {
            notprime[i * prime[j]] = 1;
            if (i % prime[j])
            {
                mu[i * prime[j]] = mu[i] * (-1);
            }
            else
            {
                mu[i * prime[j]] = 0;
                break;
            }
        }
    }
}
bool cmp(int a, int b)
{
    return a > b;
}
ll ans;
int n, v[N], a[N], dd[N], b[N];
int main()
{

    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {

        scanf("%d", &a[i]);
        v[a[i]]++;
    }
    sort(a + 1, a + 1 + n, cmp);
    for (int i = 1; i < N; i++)
        if (v[i] >= 2)
            ans = max(1ll * i, ans);
    init();
    for (int g = 1; g <= sqrt(N); g++)
    {
        stack<int> s;
        int num = 0;
        for (int i = 1; i <= n; i++)
            if (a[i] % g == 0)
                b[++num] = a[i] / g;
        if (num <= 1)
            continue;
        for (int i = 1; i <= num; i++)
        {
            //cout << g << " " << a[i] << endl;
            int res = 0, x = b[i];
            for (int t = 0; t < d[x].size(); t++)
                res += dd[d[x][t]] * mu[d[x][t]];
            //cout << res << endl;
            while (res > 0 && !s.empty())
            {
                int y = s.top();
                s.pop();
                if (gcd(x, y) == 1)
                {
                    ans = max(ans, 1ll * x * y * g);
                    res--;
                }
                for (int t = 0; t < d[y].size(); t++)
                    dd[d[y][t]]--;
            }
            s.push(x);
            for (int t = 0; t < d[x].size(); t++)
                dd[d[x][t]]++;
        }
        while (!s.empty())
        {
            int y = s.top();
            s.pop();
            for (int t = 0; t < d[y].size(); t++)
                dd[d[y][t]]--;
        }
    }
    printf("%lld\n", ans);
}