QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#788419#6620. Linear Fractional TransformationLoxilanteWA 167ms3892kbC++201.2kb2024-11-27 16:51:552024-11-27 16:51:59

Judging History

你现在查看的是最新测评结果

  • [2024-11-27 16:51:59]
  • 评测
  • 测评结果:WA
  • 用时:167ms
  • 内存:3892kb
  • [2024-11-27 16:51:55]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
typedef long double db;
const db PI = acos(-1.0);
struct Complex {
	db x, y;
	Complex(db _x = 0.0, db _y = 0.0) {
		x = _x;
		y = _y;
	}
	Complex operator - (const Complex& b) const {
		return Complex(x - b.x, y - b.y);
	}
	Complex operator + (const Complex& b) const {
		return Complex(x + b.x, y + b.y);
	}
	Complex operator * (const Complex& b) const {
		return Complex(x * b.x - y * b.y, x * b.y + y * b.x);
	}
	Complex operator / (const Complex& p) const {
		db a = x, b = y, c = p.x, d = p.y;
		return Complex((a * c + b * d) / (c * c + d * d), (b * c - a * d) / (c * c + d * d));
	}
	bool operator == (const Complex& b) const {
		return x == b.x && y == b.y;
	}
	void input() {
		scanf("%Lf%Lf", &x, &y);
	}
	void output() {
		printf("%.10Lf %.10Lf\n", x, y);
	}
}z[5], w[5];
int main() {
	int T;
	cin >> T;
	while (T--) {
		for (int i = 1; i <= 3; i++) {
			z[i].input();
			w[i].input();
		}
		z[4].input();
		Complex r = ((z[4] - z[1]) * (z[3] - z[2]) * (w[3] - w[1])) / ((z[4] - z[2]) *  (z[3] - z[1]) * (w[3] - w[2]));
		r = (w[1] - r * w[2]) / (Complex(1, 0) - r);
		r.output();
	}
	return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 3884kb

input:

2
-1 0 0 -1
0 1 -1 0
1 0 0 1
0 -1
-1 0 -1 0
0 1 0 -1
1 0 1 0
0 -1

output:

1.0000000000 0.0000000000
0.0000000000 1.0000000000

result:

ok 4 numbers

Test #2:

score: -100
Wrong Answer
time: 167ms
memory: 3892kb

input:

100000
0 0 -1 1
1 1 1 0
1 0 1 -1
-1 0
-1 -1 -1 1
1 -1 1 -1
-1 0 1 0
-1 -1
-1 -1 0 -1
-1 1 -1 -1
0 -1 0 0
1 1
1 0 0 -1
0 0 0 0
-1 -1 1 0
1 1
-1 -1 0 -1
0 1 1 -1
1 0 -1 -1
1 -1
0 1 1 -1
1 0 1 0
0 -1 0 1
-1 -1
1 1 -1 1
0 0 -1 -1
0 1 0 1
1 0
1 1 1 -1
0 1 -1 -1
1 0 -1 0
1 -1
1 0 -1 1
-1 -1 1 0
0 -1 0 1
0...

output:

1.0000000000 1.6666666667
-1.0000000000 1.0000000000
-1.5000000000 -0.5000000000
0.3333333333 -0.6666666667
-0.3846153846 -0.9230769231
-1.0000000000 -0.0000000000
-1.5000000000 0.5000000000
-0.8536585366 -0.3170731707
-1.3333333333 0.3333333333
0.1724137931 -0.0689655172
-nan -nan
-0.6666666667 -1....

result:

wrong output format Expected double, but "-nan" found