51nod 1060

反素数
定义：对于任意正整数 $n$, 其约数个数记为 $f(n)$, 如果某个正整数 $n$ 满足

对于任意正整数 $i, (0 < i < n)$, 都有 $f(i) < f(n)$，称 $n$ 为反素数。

so
一个反素数的所有因子必然是从 $2$ 开始的连续质数
如果 $n = 2 ^ {t_1} * 3 ^ {t_2} * 5 ^ {t_3} * \dots$, 那么必有 $t_1 \ge t_2 \ge t_3 \ge \dots$

结合质因数个数的统计方法即可解决此题

#include <bits/stdc++.h>

#define gc getchar()
#define LL long long

inline LL read() {
    LL x = 0; char c = gc;
    while(c < '0' || c > '9') c = gc;
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
    return  x;
}

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

LL Num, js, n;

void Dfs(LL Now_num, LL Now_js, int dep, int pre) {
    if(dep >= To) return ;
    if(Now_js > js) {
        Num = Now_num, js = Now_js;
    } else if(Now_js == js) {
        Num = std:: min(Num, Now_num);
    }
    for(int i = 1; i <= pre; i ++) {
        if(Now_num <= n / prime[dep]) {
            Now_num *= prime[dep];
            Dfs(Now_num, Now_js * (i + 1), dep + 1, i);
        } else break;
    }
}

int main() {
    int T = read();
    for(; T; T --) {
        n = read();
        Num = 1, js = 1;
        Dfs(1, 1, 0, 15);
        printf("%I64d %I64d\n", Num, js);
    }
    
    
    return 0;
}