[CF1493D] GCD of an Array - 数论,map,set

[CF1493D] GCD of an Array - 数论,map,set

Description

一个长度为 \(n\) 的序列 \(a\)。\(m\) 次操作，每次操作你要将 \(a_x\) 乘 \(y\)，然后输出 \(\gcd(a_1,\cdots,a_n)\bmod (10^9+7)\)。\(1\le n,m\le 2\times 10^5\)，\(1\le a_i,y\le 2\times 10^5\)，\(1\le x\le n\)。

Solution

对每个数用 map 维护其质因子出现次数，对每个质因子用 multiset 维护其在所有数中的出现次数

#include <bits/stdc++.h>
using namespace std;

#define int long long
const int N = 200005;
const int mod = 1e9 + 7;

map<int, int> a[N];
multiset<int> s[N];

int qpow(int p, int q)
{
    return (q & 1 ? p : 1) * (q ? qpow(p * p % mod, q / 2) : 1) % mod;
}

int inv(int p)
{
    return qpow(p, mod - 2);
}

vector<pair<int, int>> Factorial(int n)
{
    vector<pair<int, int>> ans;
    for (int i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
        {
            int cnt = 0;
            while (n % i == 0)
                n /= i, ++cnt;
            ans.push_back({i, cnt});
        }
    }
    if (n > 1)
        ans.push_back({n, 1});
    return ans;
}

signed main()
{
    ios::sync_with_stdio(false);

    int n, q;
    cin >> n >> q;

    int ans = 1;

    for (int i = 1; i <= n; i++)
    {
        int x;
        cin >> x;
        vector<pair<int, int>> fac = Factorial(x);
        for (auto [x, y] : fac)
        {
            a[i][x] += y;
            s[x].insert(y);
        }
    }

    for (int i = 1; i < N; i++)
    {
        if (s[i].size() == n)
        {
            int q = *s[i].begin();
            ans = (ans * qpow(i, q)) % mod;
        }
    }

    for (int i = 1; i <= q; i++)
    {
        int pos, val;
        cin >> pos >> val;
        vector<pair<int, int>> fac = Factorial(val);
        for (auto [x, y] : fac)
        {
            int old = a[pos][x];
            if (old && s[x].size() == n)
                ans = (ans * inv(qpow(x, *s[x].begin()))) % mod;
            if (old)
                s[x].erase(s[x].find(old));
            int tmp = a[pos][x] += y;
            s[x].insert(tmp);
            if (tmp && s[x].size() == n)
                ans = (ans * (qpow(x, *s[x].begin()))) % mod;
        }
        cout << ans << endl;
    }
}

标签：map,set,GCD,qpow,int,le,ans,mod
来源： https://www.cnblogs.com/mollnn/p/14535638.html