Looksery Cup 2015 H. Degenerate Matrix 数学

H. Degenerate Matrix

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/549/problem/H

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 matrix A is defined as a maximum of absolute values of its elements.

You are given a matrix . Consider any degenerate matrix B such that norm ||A - B|| is minimum possible. Determine ||A - B||.

Input

The first line contains two integers a and b (|a|, |b| ≤ 109), the elements of the first row of matrix A.

The second line contains two integers c and d (|c|, |d| ≤ 109) the elements of the second row of matrix A.

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 exceed 10 - 9.

Sample Input

1 2
3 4

Sample Output

0.2000000000

HINT

题意

给你一个矩阵，然后要求俩矩阵中元素相剪的最大值最小，求第二个矩阵

第二个矩阵满足ad-bc=0

题解：

2种方法可以解决这个问题

1.推公式

2.二分答案

代码：

 

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 2000001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

int main()
{
    ll a,b,c,d;
    //double a1,b1,c1,d1;
    cin>>a>>b>>c>>d;
    ll tmp=abs(a+b+c+d);
    tmp=max(tmp,abs(d+c-a-b));
    tmp=max(abs(d+b-a-c),tmp);
    tmp=max(abs(b+c-d-a),tmp);
    if(tmp==0)
        cout<<"0"<<endl;
    else
    {
        printf("%.12lf",(1.*abs(a*d-b*c)/(1.*tmp)));
    }
}