题目链接

先说一下反素数:(引自百度百科)

基本概念

定义

对于任何正整数x,其约数的个数记做g(x).例如g(1)=1,g(6)=4.如果某个正整数x满足:对于任意i(0<i<x),都有g(i)<g(x),则称x为反素数.
性质

性质一:一个反素数的质因子必然是从2开始连续的质数.
性质二:p=2^p1*3^p2*5^p3*7^p4…..必然p1>=p2>=p3>=….

这两条性质决定了搜索复杂度不会很高

想了想大概的证明:对于1-x的反素数n,可以表示为a1^p1 * a2^p2 * …. an^pn

1.假如a不连续,即存在ai 和 ai+1在素数表中不相邻,那么总可以把ai+1换成比ai大的最小的素数,仍然使得结果不变.得证

2.假如pi < pj (i < j),那么将ai与aj的个数对调,结果也不变,得到的数减少.得证

#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
#include <bitset>
#include <fstream>
using namespace std;
//LOOP
#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
//OTHER
#define SZ(V) (int)V.size()
#define PB push_back
#define MP make_pair
#define all(x) (x).begin(),(x).end()
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RS(s) scanf("%s", s)
//OUTPUT
#define WI(n) printf("%d\n", n)
#define WS(n) printf("%s\n", n)
//debug
//#define online_judge
#ifndef online_judge
#define debugt(a) cout << (#a) << "=" << a << " ";
#define debugI(a) debugt(a) cout << endl
#define debugII(a, b) debugt(a) debugt(b) cout << endl
#define debugIII(a, b, c) debugt(a) debugt(b) debugt(c) cout << endl
#define debugIV(a, b, c, d) debugt(a) debugt(b) debugt(c) debugt(d) cout << endl
#else
#define debugI(v)
#define debugII(a, b)
#define debugIII(a, b, c)
#define debugIV(a, b, c, d)
#endif
#define sqr(x) (x) * (x)
typedef long long LL;
typedef vector <int> VI;
const int INF = 0x3f3f3f3f;
const double EPS = 1e-10;
const int MOD = 100000007;
const int MAXN = 100010;
const double PI = acos(-1.0);
int prime[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 47, 53, 57};
LL ipt, dfs_n, dfs_ans;
int maxfac;
void dfs(int index, int cnt, int pre, int ans, LL mul)
{
    if (cnt == maxfac || mul > ipt / prime[index])
    {
        if (ans > dfs_ans)
        {
            dfs_ans = ans;
            dfs_n = mul;
        }
        if (ans == dfs_ans && mul < dfs_n)
        {
            dfs_n = mul;
        }
        return;
    }
    LL t = prime[index];
    FE(i, 1, min(pre, maxfac - cnt))
    {
        if (mul > ipt / t) return;
        dfs(index + 1, cnt + i, i, ans * (i + 1), mul * t);
        if (t > ipt / prime[index]) break;
        t *= prime[index];
    }
}
int main()
{
    int kase;
    RI(kase);
    while (kase--)
    {
        dfs_ans = -1;
        scanf("%I64d", &ipt);
        maxfac = (int)log2(ipt * 1.0);
        dfs(0, 0, INF, 1, 1);
        printf("%I64d %I64d\n", dfs_n, dfs_ans);
    }
    return 0;
}