实在没看懂，数组大小开的应该是够的吧，subtask3和4全WA，相当于没加强的数据都能对，加强了的都错了

#include<bits/stdc++.h>
#define maxa 100010
#define N 10000010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
inline ll read(){
	char c=getchar();ll ans=0;
	while(c<'0'||c>'9')c=getchar();
	while(c>='0'&&c<='9')ans=ans*10+c-'0',c=getchar();
	return ans;
}
ll n,u,mx,a[maxa];
ll q[N<<2],p[N<<2],lq=1,rq,lp=1,rp;
ll sq,sp,x,op,ans;
signed main(){
	n=read();
	for(int i=1;i<=n;i++){
		u=read();
		a[u]++;
		mx=max(mx,u);
	}
	for(int i=1;i<=mx;i++){
		while(a[i]--){
			q[++rq]=i;
//			printf("%lld ",q[rq]);
		}
	}
	sq=n;
	while(sq+sp>1){
		x=inf,op=0;
		if(sq>=2){
			if(q[lq]+q[lq+1]<x){
				x=q[lq]+q[lq+1];
				op=1;
			}
		}
		if(sp>=2){
			if(p[lp]+p[lp+1]<x){
				x=p[lp]+p[lp+1];
				op=2;
			}
		}
		if(sq&&sp){
			if(q[lq]+p[lp]<x){
				x=q[lq]+p[lp];
				op=3;
			}
		}
		if(op==1){
			p[++rp]=q[lq]+q[lq+1];
			lq+=2,sq-=2,sp++;
		}else if(op==2){
			p[++rp]=p[lp]+p[lp+1];
			lp+=2,sp--;
		}else{
			p[++rp]=q[lq]+p[lp];
			lq++,lp++,sq--;
		}
		ans+=p[rp];
	}
	printf("%lld\n",ans);
	return 0;
}