思路：开始没有思路，想到了用三边乱搞（每条边按照比例增加）然而样例都无法通过。后来想到了海伦公式sqrt((a+b+c)(a+b-c)(b+c-a)(a+c-b))/4，那么这样以来就是让这个三角形趋于正三角形了，即三边的方差最小，那么依次是先补最短，次短，长（贪心的算法）。

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=1000000;
const int M=2005;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define fr(i,s,n) for (int i=s;i<=n;i++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);double a[3],l;
double sd(double a,double b,double c)
{double p=(a+b+c)*0.5;return sqrt(p*(p-a)*(p-b)*(p-c));
}double sol()
{double t=min(l,a[1]-a[0]);a[0]+=t,l-=t;if (l>0){t=min(l*0.5,(a[2]-a[1]));a[1]+=t;a[0]+=t;l-=t*2;//cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<endl;if (l>0){fr(i,0,2)a[i]+=1.0*l/3;}}//cout<<a[0]<<" "<<a[1]<<" "<<a[2]<<endl;return sd(a[0],a[1],a[2]);
}
int main()
{int T;scanf("%d",&T);while (T--){scanf("%lf %lf %lf %lf",&a[0],&a[1],&a[2],&l);sort(a,a+3);printf("%.10f\n",sol());}
}