题解

从每个质因子小到大指数非严格递减，直接搜就行

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 + c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {out(x / 10);}
    putchar('0' + x % 10);
}
int N,ans,res;
int prime[MAXN],tot;
bool isprime[MAXN];
void dfs(int pos,int val,int pre,int cnt) {
    if(cnt <= ans && val > res) return;
    if(cnt > ans) {
    res = val;
    ans = cnt;
    }
    else if(cnt == ans && val < res) res = val;
    int t = 1;
    for(int i = 1 ; i <= pre ; ++i) {
    t = t * prime[pos];
    if(val > N / t) break;
    dfs(pos + 1,val * t,i,cnt * (i + 1));
    }
}
void Solve() {
    read(N);
    for(int i = 2 ; i <= 100000 ; ++i) {
    if(!isprime[i]) {
        prime[++tot] = i;
    }
    for(int j = 1 ; j <= tot ; ++j) {
        if(prime[j] > 100000 / i) break;
        isprime[i * prime[j]] = 1;
        if(i % prime[j] == 0) break;
    }
    }
    dfs(1,1,32,1);
    out(res);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}