你复杂度 $O(n^2 \log n)$，不 TLE 是什么。
```cpp
#include <bits/stdc++.h>
using namespace std;
priority_queue<int,vector<int>,greater<int> > q;
int n,x,ans = 0;
int main(){
scanf("%d",&n);
for (int i = 1; i <= n; i++) scanf("%d",&x),q.push(x);
for (int i = 2; i <= n; i++) {
int x1 = q.top();
q.pop();
int x2 = q.top();
q.pop();
int sum = x1 + x2;
ans += sum;
q.push(sum);
}
printf("%d",ans);
return 0;
}
```
我的优先队列 $43$ ms 秒杀。
by liuruiqing @ 2024-03-23 15:30:41
@[Leo_Lyc](/user/1052917) 把O2 优化打开就可以了
by lzaytyj @ 2024-03-23 15:43:52
@[lzaytyj](/user/1235091) @[liuruiqing](/user/1118614) thx!
by Leo_Lyc @ 2024-03-25 20:56:18
@[Leo_Lyc](/user/1052917) 啥意思
by lzaytyj @ 2024-03-27 22:34:08
这题优先队列是最好的做法，不接受反驳
by remake1958 @ 2024-04-03 18:27:29
@[lzaytyj](/user/1235091) thx=than - ks=thanks
x=/ks/
by Leo_Lyc @ 2024-04-06 14:31:16
@[remake1958](/user/1272803) 我也是这么做的
by Mathhew @ 2024-05-03 12:14:17