解法一：筛法求素数

筛法求素数，然后从小到大判断。

#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 100000;
vector<int> primes;
void getPrimes() {
    bool flag[MAX_N];
    fill(flag, flag + MAX_N, true);
    for (int i = 2; i <= MAX_N; ++i) {
        if (flag[i]) {
            for (int j = i << 1; j <= MAX_N; j += i) {
                flag[j] = false;
            }
            primes.emplace_back(i);
        }
    }
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    getPrimes();
    int N;
    cin >> N;
    if (N == 1) {
        cout << "1=1\n";
        return 0;
    }
    cout << N << "=";
    for (auto prime:primes) {
        if (N == 1) {
            break;
        }
        int cnt = 0;
        while (N % prime == 0) {
            cnt++;
            N /= prime;
        }
        if (cnt > 0) {
            if (cnt == 1) {
                if (N == 1) {
                    cout << prime << "\n";
                    return 0;
                } else {
                    cout << prime << "*";
                }
            } else {
                if (N == 1) {
                    cout << prime << "^" << cnt << "\n";
                    return 0;
                } else {
                    cout << prime << "^" << cnt << "*";
                }
            }
        }
    }
}