暴力出奇迹，打表过样例

现在是骗分时间

我们注意到小于2e9的反素数为：

1,2,4,6,12,24,36,48,60,120,180,240,360,720,840,1260,1680,2520,5040,7560,10080,15120,20160,25200,27720,45360,50400,55440,83160,110880,166320,221760,277200,332640,498960,554400,665280,720720,1081080,1441440,2162160,2882880,3603600,4324320,6486480,7207200,8648640,10810800,14414400,17297280,21621600,32432400,36756720,43243200,61261200,73513440,110270160,122522400,147026880,183783600,245044800,294053760,367567200,551350800,698377680,735134400,1102701600,1396755360

因此我们可以写出AC代码：

#include <bits/stdc++.h>
using namespace std;
int n,p[1145] = {0,1,2,4,6,12,24,36,48,60,120,180,240,360,720,840,1260,1680,2520,5040,7560,10080,15120,20160,25200,27720,45360,50400,55440,83160,110880,166320,221760,277200,332640,498960,554400,665280,720720,1081080,1441440,2162160,2882880,3603600,4324320,6486480,7207200,8648640,10810800,14414400,17297280,21621600,32432400,36756720,43243200,61261200,73513440,110270160,122522400,147026880,183783600,245044800,294053760,367567200,551350800,698377680,735134400,1102701600,1396755360};
int main(){
	cin >> n;
	for(int i = 1 ; i <= 68 ; i++){
		if(n < p[i]){
			cout << p[i-1] << endl;
			return 0;
		}
	}
	cout << p[68] << endl;
	return 0;
}

C++ :

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
long long p[20]= {0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};
long long maxn=-1,ans=-1;
long long n=0;
void get(long long m,long long f,long long t,long long pr) {
    //f为当前质数的编号，当前指数<pr
//t为当前约数的个数， m表示当前可能成为最优解的数
    if(t>maxn || (t==maxn && m<ans)) { //更新最优解
        ans=m,maxn=t;
    }
    long long i=m,j=0;
    long long nt=0;
    while(j<pr) { //j表示的是当前正在搜索的指数
        j++;
        if(n/i<p[f]) { //若不满足条件就跳出循环（i表示的是当前的m）
            break;
        }
        nt=t*(j+1),i*=p[f];//更新新数以及它的因子个数。
        if(i<=n) { //若i（即当前的m）在区间[1,n]内就继续搜索。
            get(i,f+1,nt,j);
        }
    }
}
int main() {
    scanf("%lld",&n);
    get(1,1,1,30);
    printf("%lld",ans);
    return 0;
}