Looksery Cup 2015-H. Degenerate Matrix(浮点数二分)

原题链接

H. Degenerate Matrix

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

input

1 23 4

output

0.2000000000

input

1 00 1

output

0.5000000000

假设矩阵的四个值为a, b, (第一行两个元素)c, d(第二行两个元素) , 二分枚举最大值mid, 那么a, b, c, d每个数都可以上下浮动mid, 算出a * d的最大值r1, 最小值l1, b * c的最大值r2, 最小值l2, 若[l1, r1]和[l2, r2]相交则可在最大值为mid之内构成一个B矩阵

#include <cstdio>#include <iostream>#define maxn 1005 #define MOD 1000000007typedef long long ll;using namespace std;int main(){ll a, b, c, d;scanf("%I64d%I64d%I64d%I64d", &a, &b, &c, &d);double l = 0, r = 2e9, a1, a2, b1, b2, c1, c2, d1, d2;for(int i = 0; i < 100000; i++){double 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;double t1 = max(max(a1 * d1, a1 * d2), max(a2 * d1, a2 * d2));double t2 = min(min(a1 * d1, a1 * d2), min(a2 * d1, a2 * d2));double t3 = max(max(b1 * c1, b1 * c2), max(b2 * c1, b2 * c2));double t4 = min(min(b1 * c1, b1 * c2), min(b2 * c1, b2 * c2));if(t4 <= t1 && t2 <= t3) r = mid;else l = mid;}printf("%.9lf\n", l);return 0;}