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Thwrani
Asked:2022-08-17 14:50:10 +0000 UTC2022-08-17 14:50:10 +0000 UTC

任务“学位”

  • 772

给定任务：

为了检查她的学生如何数数，Ma​​ria Ivanovna 每年都会在家里给他们布置相同的任务 - 对于给定的自然 A，找到最小自然 N 使得 N 的 N 次方（N 乘以自身 N 次）是能被 A 整除。有必要写一个程序来解决这个问题

1 ≤ A ≤ 10^9
我写了这段代码：

#include <iostream>
#include <vector>
#include <cmath>
#include <unordered_set>
using namespace std;
vector<int> decomp (int n)
{
    vector<int> ans (0);
    int d = 2;
    while (d * d <= n)
    {
        if (n % d == 0)
        {
            ans.push_back(d);
            n /= d;
        }
        else
            d += 1;
    }
    if (n > 1)
        ans.push_back(n);
    return ans;
}
int main()
{
    int x; cin >> x;
    if (x == 1)
        cout << 1;
    else
    {
        vector<int> tmp = decomp(x);
        unordered_set<int> s;
        for (int el: tmp)
            s.insert(el);
        vector<int> a (0);
        for (int el: s)
            a.push_back(el);
        int y = 1;
        for (int i = 0; i < a.size(); i++)
            y *= a[i];
        if (y >= 29)
            cout << y;
        else
        {
            int k = 1; int n = k * y;
            while ((int)pow(n, n) % x != 0)
            {
                n = k * y;
                k += 1;
            }
            cout << n;
        }
    }
}

但是它运行得太慢了。请帮助我，我该如何优化它？

c++ оптимизация

1 个回答

  1. Best Answer

    好吧，例如，像这样：

    #include <iostream>
#include <map>

using namespace std;

map<unsigned int, unsigned int> factor(unsigned int A)
{
    map<unsigned int,unsigned int> m;
    while(A%2 == 0) { m[2]++; A /= 2; };
    for(unsigned int i = 3; i*i <= A; i += 2)
    {
        while(A%i == 0) { m[i]++; A /= i; };
    }
    if (A != 1) m[A]++;
    return m;
}

int main(int argc, const char * argv[])
{
    unsigned int A;
    cin >> A;
    if (A == 1) { cout << "1\n"; return 0; }
    auto m = factor(A);
    unsigned int p = 1;
    unsigned int e = 0;
    for(const auto& v: m)
    {
        p *= v.first;
        if (v.second > e) e = v.second;
        //cout << v.first << " ** " << v.second << endl;
    }
    if (p < e)
    {
        unsigned int N = ((e+p-1)/p)*p;
        unsigned int M = p, k = 1;

        while(M*k < e)
        {
            for(const auto& v: m)
            {
                if (v.second == e)
                {
                    M *= v.first;
                    k++;
                }
            }
        }
        cout << ((M < N) ? M : N) << endl;
    }
    else cout << p << endl;
}
    • 3

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