二分法

题意：奇异矩阵 使范式最大的最小

#include <iostream>
#include <stdio.h>
#include <string.h>
#include<algorithm>
#include<cmath>
using namespace std;
double a, b, c, d;
double max1;
double min1;
double max2;
double min2;
void fun1(double a1,double a2,double b1,double b2)
{
    double q[4];
    q[0] = a1*b1;
    q[1] = a1*b2;
    q[2] = a2*b1;
    q[3] = a2*b2;
    sort(q, q + 4);
    max1 = q[3];
    min1 = q[0];
}
void fun2(double a1, double a2, double b1, double b2)
{
    double q[4];
    q[0] = a1*b1;
    q[1] = a1*b2;
    q[2] = a2*b1;
    q[3] = a2*b2;
    sort(q, q + 4);
    max2 = q[3];
    min2 = q[0];
}
bool cmp(double a1, double a2, double a3)
{
    if (a3 >= a1&&a3 <= a2)
        return true;//在范围内
    return false;
}
bool check(double n)
{
    double a1 = a + n;
    double a2 = a - n;
    double b1 = b + n;
    double b2 = b - n;
    double c1 = c + n;
    double c2 = c - n;
    double d1 = d + n;
    double d2 = d - n;
    fun1( a1, a2, d1, d2);
    fun2(b1, b2, c1, c2);
    if (cmp(min1, max1, min2) || cmp(min2,max2,min1))
    return true;
    return false;
}
int main()
{
    scanf("%lf%lf%lf%lf", &a, &b, &c, &d);
    double m=-2e10;
    if (m < a)
    {
        m = a;
    }
    if (m < b)
    {
        m = b;
    }
    if (m < c)
    {
        m = c;
    }
    if (m <d)
    {
        m = d;
    }
    double left, right;
    left = 0;
    right = m;
    double mid;
    while (fabs(left -right)>0.0000001)//求的是最大里面最小的
    {
        mid = (left + right) / 2.0;
        if (check(mid))//满足条件
        {
            right = mid;//大了
        }
        else
        {
            left = mid + 0.0000005;//小了
        }
    }
    printf("%lf\n", right);
    return 0;
}