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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
#244902#6620. Linear Fractional TransformationariesTL 0ms3796kbC++143.2kb2023-11-09 16:53:582023-11-09 16:53:58

Judging History

你现在查看的是最新测评结果

  • [2023-11-09 16:53:58]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:3796kb
  • [2023-11-09 16:53:58]
  • 提交

answer

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

typedef long long ll;

char *p1,*p2,buf[100000];
#define nc() (p1==p2 && (p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++)
int read()
{
    int x=0,f=1;
    char ch=nc();
    while(ch<48||ch>57)
    {
        if(ch=='-')
            f=-1;
        ch=nc();
    }
    while(ch>=48&&ch<=57)
        x=x*10+ch-48,ch=nc();
   	return x*f;
}

struct com{
    double x,y;  
    friend com operator +(const com&a,const com&b){
        com c;
        c.x = a.x + b.x;
        c.y = a.y + b.y;
        return c;
    }
    friend com operator -(const com&a,const com&b){
        com c;
        c.x = a.x - b.x;
        c.y = a.y - b.y;
        return c;
    }
    friend com operator *(const com&a,const com&b){
        com c;
        c.x = a.x*b.x - a.y*b.y;
        c.y = a.y*b.x + a.x*b.y;
        return c;
    }
    friend com operator /(const com&a,const com&b){
        com c;
        c.x = (a.x*b.x + a.y*b.y)/(b.x*b.x + b.y*b.y);
        c.y = (-a.x*b.y+a.y*b.x)/(b.x*b.x + b.y*b.y);
        return c;
    }
    
    friend bool operator ==(const com&a,const com&b){
        return fabs(a.x-b.x)<1e-7 && fabs(a.y-b.y)<1e-7; 
    }

    friend com operator -(const com&a){
        com c;
        c.x = -a.x;c.y = -a.y;
        return c; 
    }

};

com m[10][10];


com fun(int a,int b,int c)
{
    return m[1][a]*(m[2][b]*m[3][c] - m[2][c]*m[3][b]) + m[1][b]*(m[2][c]*m[3][a] - m[2][a]*m[3][c])
        + m[1][c]*(m[2][a]*m[3][b] - m[2][b]*m[3][a]);
}

void solve()
{
    com z1,z2,z3,w1,w2,w3,z0,w0;
    z1.x =read();z1.y = read();w1.x = read();w1.y=read();
    z2.x =read();z2.y = read();w2.x = read();w2.y=read();
    z3.x =read();z3.y = read();w3.x = read();w3.y=read();
    z0.x =read();z0.y = read();
    // cin>>z1.x>>z1.y>>w1.x>>w1.y;
    // cin>>z2.x>>z2.y>>w2.x>>w2.y;
    // cin>>z3.x>>z3.y>>w3.x>>w3.y;
    // cin>>z0.x>>z0.y;
    
    com a,b,c,d;

    //d=0 c!=0  ->  c=1  // (az+b)/z 此时不能有z为0


    a = (w1*z1 - w2*z2)/(z1-z2);
    b = w1*z1 - a*z1;

        
    if((a*z3 + b)/z3== w3){
        w0 = (a*z0+b)/z0;
        cout<<w0.x<< " "<<w0.y<<"\n";
        return;
    }

    //az+b/cz+d = w
    //az + b - czw = w;

    //d!=0 -> d=1
    d.x=1;d.y=0;
    
    //c=0
    a = (w1-w2)/(z1-z2);
    b = w1-a*z1;
    if((a*z3 + b) == w3){
        w0 = a*z0+b;
        cout<<w0.x<< " "<<w0.y<<"\n";
        return;
    }

    //c!=0

    m[1][1] = z1;m[1][2] = d;m[1][3] = -z1*w1;m[1][4] = w1;
    m[2][1] = z2;m[2][2] = d;m[2][3] = -z2*w2;m[2][4] = w2;
    m[3][1] = z3;m[3][2] = d;m[3][3] = -z3*w3;m[3][4] = w3;
    
    com t,ta,tb,tc;
    t = fun(1,2,3);

    ta = fun(4,2,3);
    tb = fun(1,4,3);
    tc = fun(1,2,4);

    a = ta/t;
    b = tb/t;
    c = tc/t;
    
    w0 = (a*z0 + b)/(c*z0 + d);
    cout<<w0.x<<" "<<w0.y<<"\n";

}




int main()
{
    // ios::sync_with_stdio(0);
    // cin.tie(0);cout.tie(0);
    ll T=read();
    cin>>T;

    cout<<fixed<<setprecision(15);

    while(T--){
        solve();
    }
    return 0;
}


/*

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




*/

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3796kb

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:

1.000000000000000 0.000000000000000
0.000000000000000 1.000000000000000

result:

ok 4 numbers

Test #2:

score: -100
Time Limit Exceeded

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:

1.000000000000000 1.666666666666667
-1.000000000000000 1.000000000000000
-1.500000000000000 -0.500000000000000
0.333333333333333 -0.666666666666667
-0.384615384615385 -0.923076923076923
-1.000000000000000 0.000000000000000
-1.500000000000000 0.500000000000000
-0.853658536585366 -0.317073170731707
-1...

result: