Linear Fractional Transformation

NC230864. Linear Fractional Transformation

描述

The linear fractional transformations are the functions $f(z)=\frac{az+b}{cz+d}$ $(a,b,c,d \in \mathbb{C}, ad-bc \neq 0)$ mapping the extended complex plane $\mathbb{C} \cup \{ \infty \}$ onto itself.

Given $f(z_1) = w_1$ , $f(z_2) = w_2$ and $f(z_3) = w_3$ , where $z_1$ , $z_2$ and $z_3$ are pairwise distinct complex numbers and $w_1$ , $w_2$ and $w_3$ are also pairwise distinct complex numbers, your task is to calculate $f(z_0)$ for a certain complex number $z_0$ . It can be shown that the answer is always unique to the given contraints.

输入描述

The input contains several test cases, and the first line contains an integer  , indicating the number of test cases.

To clarify the input format, we denote , , , , ,  and , where  is the imaginary unit that .

Then for each test case, the first line contains four integers , ,  and , the second contains four integers , ,  and , the third line contains four integers , ,  and , and the fourth line contains only two integers  and . It is guaranteed that all these integers are in the range  and the answer  satisfies , where   is the modulus of the complex number .

输出描述

For each test case, output a line containing two real numbers  and , indicating the real part and the imaginary part of .

Your answer is acceptable if the absolute or relative errors of both the real part and the imaginary part do not exceed . Formally speaking, suppose that your output is  and the jury's answer is , your output is accepted if and only if .

示例1

输入:

输出:

1.000000000000000 0.000000000000000
0.000000000000000 1.000000000000000

说明:

In the first sample case we have $f(z)=iz$ , and in the second sample case we have $f(z)=1/z$ .

原站题解

上次编辑到这里，代码来自缓存点击恢复默认模板

C++ 解法, 执行用时: 528ms, 内存消耗: 8320K, 提交时间: 2021-11-23 19:47:43

#include<bits/stdc++.h>
using namespace std;

typedef complex<double> C;

C getC(){
    int x,y;cin>>x>>y;
    return C(x,y);
}
int T;
int main(){
    int t;scanf("%d",&t);
    while(t--){
        C z_0=getC(),w_0=getC(),z_1=getC(),w_1=getC(),z_2=getC(),w_2=getC(),z_3=getC();
        C t=(z_0-z_3)/(z_1-z_3)/((z_0-z_2)/(z_1-z_2))*((w_0-w_2)/(w_1-w_2))-C(1,0);
        C w_3=w_1-(w_0-w_1)/t;
        printf("%.12f %.12f\n",w_3.real(),w_3.imag());
    }
    
}

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