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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#117616#6620. Linear Fractional TransformationlsroiWA 1664ms3664kbC++142.1kb2023-07-01 20:25:452023-07-01 20:25:47

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-07-01 20:25:47]
  • 评测
  • 测评结果:WA
  • 用时:1664ms
  • 内存:3664kb
  • [2023-07-01 20:25:45]
  • 提交

answer

/*
a = (a1,a2)

*/
#include<bits/stdc++.h>
using namespace std;
const int S = 2;
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T;
	cin >> T;
	while (T--) {
		int a1, a2, b1, b2, c1, c2, d1, d2;// [-100, 100] int ?
		int r1, s1, r2, s2, r3, s3;
		int p1, q1, p2, q2, p3, q3;
		cin >> p1 >> q1 >> r1 >> s1;
		cin >> p2 >> q2 >> r2 >> s2;
		cin >> p3 >> q3 >> r3 >> s3;
		
		bool flag = false;
		for (int i = -S; i <= S; i++) {//枚举a1,a2
			for (int j = -S; j <= S; j++) {
				for (int k = -S; k <= S; k++) {//枚举c1,c2
					for (int l = -S; l <= S; l++) {
						for (int m = -S; m <= S; m++) {//枚举d1,d2
							for (int n = -S; n <= S; n++) {
								int o1 = p1*(r1*k-s1*l-i) - q1*(r1*l+s1*k-j) + (r1*m-s1*n);
								int o2 = p2*(r2*k-s2*l-i) - q2*(r2*l+s2*k-j) + (r2*m-s2*n);
								int o3 = p3*(r3*k-s3*l-i) - q3*(r3*l+s3*k-j) + (r3*m-s3*n);
								
								int u1 = p1*(r1*l+s1*k-j) + q1*(r1*k-s1*l-i) + (r1*n+s1*m);
								int u2 = p2*(r2*l+s2*k-j) + q2*(r2*k-s2*l-i) + (r2*n+s2*m);
								int u3 = p3*(r3*l+s3*k-j) + q3*(r3*k-s3*l-i) + (r3*n+s3*m);
								
								if (o1 == o2 && o1 == o3 && u1 == u2 && u1 == u3) {
									a1 = i; a2 = j;
									b1 = o1; b2 = u1;
									c1 = k; c2 = l;
									d1 = m; d2 = n;
									if (a1*d1-a2*d2 == b1*c1-b2*c2) continue;
									flag = true;
									break;
								}
							}if (flag) break;
						}if (flag) break;
					}if (flag) break;
				}if (flag) break;
			}if (flag) break;
		}
//		cout << flag << "---\n";
//		cout << a1 << ' ' << a2 << "\n";
//		cout << b1 << ' ' << b2 << "\n";
//		cout << c1 << ' ' << c2 << "\n";
//		cout << d1 << ' ' << d2 << "\n";
		
		int p, q, r, s;
		cin >> p >> q;
		
		
//		(a,b)/(c,d)=(r,s)
		int a, b, c, d;
		a = a1*p-a2*q+b1;
		b = a1*q+a2*p+b2;
		c = c1*p-c2*q+d1;
		d = c1*q+c2*p+d2;
//		cout << a << ' ' << b << ' ' << c << ' ' << d << "\n";
		
		r = a * c + b * d;
		s = b * c - a * d;
		
		cout << fixed << setprecision(10) << (double)r/(c*c+d*d) << ' ' << (double)s/(c*c+d*d) << "\n";
	}
	return 0;
}

Details

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Test #1:

score: 100
Accepted
time: 1ms
memory: 3648kb

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: 1664ms
memory: 3664kb

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:

-nan -nan
-nan -nan
-nan -nan
0.3333333333 -0.6666666667
-nan -nan
-1.0000000000 0.0000000000
-1.5000000000 0.5000000000
-nan -nan
-nan -nan
-nan -nan
-nan -nan
-nan -nan
1.0000000000 0.0000000000
0.0000000000 2.0000000000
1.5000000000 0.5000000000
2.0000000000 0.0000000000
-1.0000000000 -1.00000000...

result:

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