#include<iostream>#include<cmath>#include<iomanip>using namespace std;typedef class Node{public:double x,y;}location;double det(double x1,double y1,double x2,double y2){return x1*y2-x2*y1;}double cross(location A,location B,location C,location D) //计算 AB x CD{return det(B.x-A.x , B.y-A.y , D.x-C.x , D.y-C.y);}double xx,yy; //坐标返回值void intersection(location A,location B,location C,location D){double area1=cross(A,B,A,C);double area2=cross(A,B,A,D);xx=(area2*C.x - area1*D.x)/(area2-area1); //本题所求的交点一定是规范相交所得，因此无需判断是否规范相交yy=(area2*C.y - area1*D.y)/(area2-area1); return;}double area(location A,location B,location C,location D){double ***1=fabs(0.5*cross(A,B,A,C)); //用计算几何的方法计算的面积是有向面积double ***2=fabs(0.5*cross(A,B,A,D)); //即算出来的面积可能为负数，因此必须用绝对值 // fabs() 为取double的绝对值函数return ***1+***2;}int main(int i,int j,int k){int n;while(cin>>n){if(!n)break;location** node=new location*[n+2];node[0]=new location[n+2]; //下边node[n+1]=new location[n+2]; //上边node[0][0].x = node[0][0].y =0.0;for(i=1;i<=n;i++){cin>>node[0][i].x;node[0][i].y=0.0;}node[0][n+1].x=1.0;node[0][n+1].y=0.0;node[n+1][0].x=0.0;node[n+1][0].y=1.0;for(i=1;i<=n;i++){cin>>node[n+1][i].x;node[n+1][i].y=1.0;}node[n+1][n+1].x=1.0;node[n+1][n+1].y=1.0;for(i=1;i<=n;i++){node[i]=new location[n+2];cin>>node[i][0].y;node[i][0].x=0.0;}for(i=1;i<=n;i++){cin>>node[i][n+1].y;node[i][n+1].x=1.0;}for(j=1;j<=n;j++)for(i=1;i<=n;i++){intersection(node[0][j],node[n+1][j],node[i][0],node[i][n+1]);node[i][j].x=xx;node[i][j].y=yy;}double max_area=0.0;for(i=1;i<=n+1;i++)for(j=1;j<=n+1;j++){double temp=area(node[i-1][j-1],node[i][j],node[i][j-1],node[i-1][j]);if(max_area < temp)max_area = temp;}cout<<fixed<<setprecision(6)<<max_area<<endl;delete[] node;}return 0;}
