莫名其妙对了

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
bitset<100000010>f;
bool b;
ll prime[6000010],cnt,sig[6000010],k[6000010],maxn,hcn[6000010],tot;//sig存约数个数，k是分解质因数后当前的最小质因数的次幂
int main(){
	hcn[++tot]=1;
	maxn=1;
	for(ll i=2;i<=6000000;i++){
		if(!f[i]){
			prime[++cnt]=i;
			sig[i]=2;
			k[i]=1;
		}
		for(ll j=1,w=i*prime[j];j<=cnt&&w<=6000000;j++,w=i*prime[j]){
			f[w]=true;
			if(i%prime[j]){
				sig[w]=sig[i]*2;
				k[w]=1;
			}else{
				sig[w]=sig[i]/(k[i]+1)*(k[i]+2);
				k[w]=k[i]+1;
				break;
			}
		}
		if(sig[i]>maxn){
			maxn=sig[i];
			hcn[++tot]=i;
		}
	}//改动后的欧拉筛
	for(ll i=1;i<=tot;i++)
		printf("%lld\n",hcn[i]);//输出
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
ll n;
int main(){
	while(true){
		scanf("%lld",&n);
		for(ll i=2;i<=n;i++){
			while(n%i==0){
				printf("%lld ",i);
				n/=i;
			}
		}
		printf("\n");
	}
	return 0;
}

1

2
2
4
2 2
6
2 3
12
2 2 3
24
2 2 2 3
36
2 2 3 3
48
2 2 2 2 3
60
2 2 3 5
120
2 2 2 3 5
180
2 2 3 3 5
240
2 2 2 2 3 5
360
2 2 2 3 3 5
4324320
2 2 2 2 2 3 3 3 5 7 11 13

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
inline void in(ll &x){
	x=0;
	bool f=1;
	char c=getchar();
	for(;!isdigit(c);c=getchar())if(c=='-')f=0;
	for(;isdigit(c);c=getchar())x=x*10+(c^48);
	x=f?x:-x;
}
inline void out(ll x){
	if(x<0){putchar('-');x=-x;}
	char buf[20];
	int pos=0;
	if(x==0){putchar('0');return;}
	for(;x;x/=10)buf[pos++]=(x%10)^48;
	while(pos)putchar(buf[--pos]);
}
inline void outel(ll x){
	out(x);putchar('\n');
}//快读快写
ll prime[110]={0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541};//素数打表
ll cnt=1,n,k[100010][110],maxn,ans_cnt,ans[100010],val_maxn;//给每个反素数打个编号，ans存储反素数，k[i][j]为编号为i的反素数第j个素数的次幂，方便计算约数个数
bool f[110];
map<ll,ll>map_ans;//算是离散化存储已经加入队列的数，防止多次加入
struct node{
	ll val,w,num,ord;//val为值，w为约数个数，num为由这个编号的反素数乘上一个素数得来，ord为乘的哪个编号的素数
};
bool operator < (const node &a,const node &b){
	return a.w*b.val<b.w*a.val;//比较操作
}
priority_queue<node>q;
int main(){
	in(n);
	q.push({2,2,1,1});//将第二个反素数2加入堆顶
	ans[++ans_cnt]=1;
	map_ans[2]=1;//反素数加入一
	while(q.top().val<=n){
		node now=q.top();
		q.pop();//取出当前最优解
		if(now.val<=val_maxn||now.w<=maxn)continue;//判断当前的数是否可能成为下一个反素数
		val_maxn=now.val;
		maxn=now.w;
		ans[++ans_cnt]=now.val;
		for(ll i=1;i<=cnt;i++){
			k[ans_cnt][i]=k[now.num][i];
		}//复制一遍素数的次幂
		k[ans_cnt][now.ord]++;
		if(!f[now.ord]){
			f[now.ord]=true;
			cnt++;
		}//cnt为当前最大的反素数是哪几个不同素数的乘积
		for(ll i=1,j=now.val*prime[i];i<=cnt;i++,j=now.val*prime[i]){
			if(!map_ans.count(j)){
				q.push({j,now.w/(k[ans_cnt][i]+1)*(k[ans_cnt][i]+2),ans_cnt,i});
				map_ans[j]=1;
			}
		}//枚举乘哪个素数，将可能的答案加入优先队列，打上标记
	}//最优解在范围内就一直枚举
	outel(ans[ans_cnt]);//输出答案
	return 0;
}