codeforces 27E . Number With The Given Amount Of Divisors 搜索+数论

题目链接

首先要知道一个性质， 一个数x的因子个数等于 a1^p1 * a2^p2*....an^pn， ai是x质因子， p是质因子的个数。

然后就可以搜了

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
int n, f[105], prime[20], cnt;
ll ans = 1e18+2;
void dfs(int pos, int num, ll val) {
    if(pos>16)
        return ;
    if(num>n)
        return ;
    if(num == n) {
        ans = min(ans, val);
        return ;
    }
    for(int i = 1; i<=60; i++) {
        val *= prime[pos];
        if(val>ans)
            break;
        dfs(pos+1, num*(i+1), val);
    }
}
int main()
{
    cin>>n;
    for(int i = 2; i<100; i++) {
        if(!f[i]) {
            for(int j = i+i; j<100; j+=i) {
                f[j] = 1;
            }
            prime[cnt++] = i;
        }
    }
    dfs(0, 1, 1);
    cout<<ans<<endl;
    return 0;
}