ACM模版

求最小的因子个数为n个正整数

typedef unsigned long long ULL;

const ULL INF = ~0ULL;
const int MAXP = 16;

int prime[MAXP] = {
  2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53};

int n;
ULL ans;

void dfs(int dept, ULL tmp, int num, int pre)    // 深度/当前值/约数个数/上一个数
{
    if (num > n)
    {
        return;
    }
    if (num == n && ans > tmp)
    {
        ans = tmp;
    }
    for (int i = 1; i <= pre; i++)
    {
        if (ans / prime[dept] < tmp)
        {
            break;
        }
        dfs(dept + 1, tmp *= prime[dept], num * (i + 1), i);
    }
}

int main()
{
    while (cin >> n)
    {
        ans = INF;
        dfs(0, 1, 1, 15);
        cout << ans << endl;
    }
    return 0;
}

求n以内的因子最多的数（不止一个则取最小）

typedef long long ll;

const int MAXP = 16;
const int prime[MAXP] = {
  2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53};

ll n, res, ans;

void dfs(ll cur, ll num, int key, ll pre)  // 当前值/当前约数数量/当前深度/上一个数
{
    if (key >= MAXP)
    {
        return ;
    }
    else
    {
        if (num > ans)
        {
            res = cur;
            ans = num;
        }
        else if (num == ans)    // 如果约数数量相同，则取较小的数
        {
            res = min(cur, res);
        }

        ll i;
        for ( i = 1; i <= pre; i++)
        {
            if (cur <= n / prime[key])  // cur*prime[key]<=n
            {
                cur *= prime[key];
                dfs(cur, num * (i + 1), key + 1, i);
            }
            else
            {
                break;
            }
        }
    }
}

void solve()
{
    res = 1;
    ans = 1;

    dfs(1, 1, 0, 15);
    cout << res << ' ' << ans << endl;
}

int main(int argc, const char * argv[])
{
    int T;
    cin >> T;

    while (T--)
    {
        cin >> n;
        solve();
    }
    return 0;
}