日常水題
貪心策略：每次找代價最小的兩對合並，用小根堆來維護，每次合併之後將當前合併結果重新推入佇列，直到合併完成（n-1次）

程式碼如下

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <ctime>
#include <queue>

using namespace std;
typedef long long LL;

LL n;
LL tot,ans;
priority_queue<LL ,vector
 
<LL >,greater<LL > >q;

int main()
{
    scanf("%lld",&n);
    for(LL i = 1;i <= n;i++)
    {
        LL aa;
        scanf("%lld",&aa);
        q.push(aa);
    }
    while(q.size() != 1)
    {
        LL u = q.top();
            q.pop();
        LL v = q.top();
            q.pop();
        //cout<<"tot_after : "<<tot<<endl;
 

        tot = u+v;
        //cout<<"tot_fin : "<<tot<<endl;
        ans += tot;
        q.push(tot);
        //cout<<"ans : "<<ans<<endl;
    }
    printf("%lld\n",ans);

return 0;
}

THE END

By Peacefuldoge