題意：貪心任意石子合併

思路：平衡堆的模板和程式碼測試

#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck(x) cout<<"["<<x<<"]"
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int>PII;

const int MX = 1e3 + 5;

int rear;
LL S[MX << 2];

void push(LL x) {
    int now = ++rear, pre = now >> 1;
    S[now] = x;
    while(pre && S[pre] > S[now]) {
        swap(S[now], S[pre]);
        now = pre, pre = now >> 1;
    }
}

void pop() {
    S[1] = S[rear--];

    int now = 1;
    while((now << 1) <= rear) {
        int lt = now << 1, rt = now << 1 | 1;
        if(rt <= rear) {
            if(S[lt] >= S[now] && S[rt] >= S[now]) break;
            if(S[now] >= S[lt] && S[rt] >= S[lt]) swap(S[now], S[lt]), now = lt;
            else swap(S[now], S[rt]), now = rt;
        } else {
            if(S[lt] > S[now]) break;
            swap(S[now], S[lt]), now = lt;
        }
    }
}

int main() {
    int T, n; //FIN;
    scanf("%d", &T);
    while(T--) {
        rear = 0;
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            int t; scanf("%d", &t);
            push(t);
        }

        LL ans = 0, sum;
        for(int i = 1; i <= n - 1; i++) {
            sum = S[1]; pop();
            sum += S[1]; pop();
            ans += sum;
            push(sum);
        }
        printf("%lld\n", ans);
    }
    return 0;
}