codeforces 549H Degenerate Matrix（二分）

Description

The determinant of a matrix
2 × 2 is defined as follows:



A matrix is called degenerate if its determinant is equal to zero.

The norm||A|| of a matrixA is defined as a maximum of absolute values of its elements.

You are given a matrix

. Consider any degenerate matrixB such that norm||A - B||
is minimum possible. Determine||A - B||.

Input

The first line contains two integers a andb (|a|, |b| ≤ 109), the elements of the first row of matrixA.

The second line contains two integers c andd (|c|, |d| ≤ 109) the elements of the second row of matrixA.

Output

Output a single real number, the minimum possible value of
||A - B||. Your answer is considered to be correct if its absolute or relative error does not exceed10 - 9.

Sample Input

Input

1 2
3 4

Output

0.2000000000

Input

1 0
0 1

Output

0.5000000000

Hint

In the first sample matrix B is


In the second sample matrix B is


题目大意：定义||A||为矩阵A中的数的绝对值最大的那个值。要求找到一个矩阵B，使得||A-B||的值最小并且输出这个值。

我们可以假设要求的值最大为t。可以对差值进行二分查找。

#include<stdio.h>
#include<math.h>
#include<string.h>
double a,b,c,d;
double min(double a,double b)
{
if(a>b)return b;
else return a;
}
double max(double a,double b)
{
if(a>b)return a;
else return b;
}
int main()
{
int i,j,n,k;
double t1,t2,x1,x2;
double a1,a2,b1,b2,c1,c2,d1,d2;
while(scanf("%lf%lf%lf%lf",&a,&b,&c,&d)!=EOF)
{
double t;

double l=0,r=1000000000,mid;
for(i=0;i<100000;i++)
{
mid=(l+r)/2;
a1=a+mid;a2=a-mid;b1=b+mid;b2=b-mid;
c1=c+mid;c2=c-mid;d1=d+mid;d2=d-mid;
t1=min(min(a1*d1,a1*d2),min(a2*d1,a2*d2));
t2=min(min(b1*c1,b1*c2),min(b2*c1,b2*c2));
x1=max(max(a1*d1,a1*d2),max(a2*d1,a2*d2));
x2=max(max(b1*c1,b1*c2),max(b2*c1,b2*c2));
if(t1<=x2&&t2<=x1)r=mid;    //如果条件成立，表明我搜索的值还太大，需要缩小范围。
 else l=mid;     //表示搜索的值已经太小了
}
printf("%.11f\n",l);
}
return 0;
}