数论 CF27E Number With The Given Amount Of Divisors

求因子数一定的最小数(反素数)

 

#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#include<queue>
#include<stack>
#include<list>
#include<sstream>
#include<cstdio>
#define INF 0x3f3f3f3f
const int maxn = 1e3 + 5;
const double PI = acos(-1.0);
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

//若取前17个素数，其乘积大于要求范围
ull p[16] = { 2,3,5,7,11,13,17,23,29,31,37,41,43,47,53 };

ull ans;
ull n;

//depth 当前在枚举第几个素数，num:当前因子数
//tmp:当前因子数量为num的时候的数值
//up:上一个素数的幂，这次应该小于等于这个幂次

void dfs(ull depth, ull tmp, ull num, ull up) {
    if (num > n || depth >= 16) return;
    if (num == n && ans >= tmp) {
        ans = tmp;
        return;
    }
    for (int i = 1; i <= up; i++) {
        if (tmp / p[depth] > ans) return;
        dfs(depth + 1, tmp = tmp * p[depth], num * (i + 1), i);
    }
}

int main() {
    while (scanf("%llu", &n) != EOF) {
        ans = INF;
        dfs(0, 1, 1, 64);
        printf("%llu", ans);
    }
    return 0;
}