很简单的思路,不难看出贪心的来做,我们要从小到大排好序之后,如果有负数就让所有其他数字都加上这个数,然后继续看下一个数是否是负数 如果所有数字都会变成负数,那么最后累加的结果就是我们要的答案 如果不是所有数字都会变成负数,那么最后我们要将剩下的所有数字加上前面负数的总和的结果求和

详细代码如下:

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// GXB1  .: UBMQBBg       JB6pKpPG6O6G6EZEpEPp6MGSwPEDRQgMBP  :RPXUX6PQa;
// gXOR .,,pgGE6ODBB,      QQOZ6RMBBBQMgDZZ6ZGE21HOEZpEpEZgB,  :ZZppBQs,v
// gwXBr,;EBEEpGZGGBQ:     .RMgK7r:::i7s1HPZHDHLEGPpKpK6PpZBs .D;g6GMZ  ;

#include<bits/stdc++.h>

using namespace std;

#define int long long
#define ll long long
#define lowbit(x) x&(-x)

const int MOD=1e9 + 7;
const int N=1e6+10;

int qpow(int num,int k) {
    int res=1;
    while(k) {
        if(k%2) res = (res * num) % MOD;
        num = (num * num) % MOD;
        k/=2 ;
    }
    return res % MOD;
}

int inv(int x) {
    return qpow(x,MOD-2);
}
 
int lcm(int x,int y) {
    return x * y / __gcd(x,y);
}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int Base = uniform_int_distribution<>(8e8,9e8)(rng); 

int a[N];
int b[N];

void solve() {
    int n;
    cin>>n;
    
    priority_queue<int,vector<int>,greater<int>>q;
    for(int i=1;i<=n;i++) cin>>a[i],q.push(a[i]);
    int sum = 0;
    
    while(!q.empty())
    {
        //cout<<q.top()<<" "<<sum<<"\n";
        if(q.top() + sum <= 0)
        {
            if(q.size() == 1){
                sum = q.top() + sum;
                q.pop();
                break;
            }
            sum += q.top() + sum;
            q.pop();
        }
        else break;
    }

    int res = 0;
    while(!q.empty()){
        res += q.top() + sum;
        q.pop();
    }

    if(!res) cout<<sum<<"\n";
    else cout<<res<<"\n";
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    int t=1;
    cin>>t;

    while(t--) {
        solve();
    }
    return 0;
}