2017 ACMICPC Asia Regional Qingdao Online 1001 Apple

Time Limit: 1000/1000 MS(Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

Apple is Taotao's favouritefruit. In his backyard, there are three apple trees with coordinates (x1,y1), (x2,y2), and (x3,y3). Now Taotao is planning toplant a new one, but he is not willing to take these trees too close. Hebelieves that the new apple tree should be outside the circle which the threeapple trees that already exist is on. Taotao picked a potential position (x,y) of the new tree. Couldyou tell him if it is outside the circle or not?

 

 

Input

The first line contains aninteger T, indicating that there are T(T≤30) cases.
In the first line of each case, there are eight integers x1,y1,x2,y2,x3,y3,x,y, as described above.
The absolute values of integers in input are less than or equal to 1,000,000,000,000.
It is guaranteed that, any three of the four positions do not lie on a straightline.

 

 

Output

For each case, output"Accepted" if the position is outside the circle, or"Rejected" if the position is on or inside the circle.

 

 

Sample Input

3

-2 0 0 -2 2 0 2 -2

-2 0 0 -2 2 0 0 2

-2 0 0 -2 2 0 1 1

 

 

Sample Output

Accepted

Rejected

Rejected

最终代码还是wa

#include <iostream>#include <cmath>#include <cstdio>using namespace std;typedef long long ll;int main(){    int T;    scanf("%d",&T);    while(T--)    {        double x1,y1,x[5],y[5];        double a,b,r;        int i;        for(i=1;i<=3;i++)            scanf("%lf%lf",&x[i],&y[i]);        scanf("%lf%lf",&x1,&y1);        a = (x[2]*x[2]*y[1]-x[3]*x[3]*y[1]-x[1]*x[1]*y[2]+x[3]*x[3]*y[2]-y[1]*y[1]*y[2]+y[1]*y[2]*y[2]+x[1]*x[1]*y[3]-x[2]*x[2]*y[3]+y[1]*y[1]*y[3]-y[2]*y[2]*y[3]-y[1]*y[3]*y[3]+y[2]*y[3]*y[3])/(2* (x[2]* y[1] - x[3] *y[1] - x[1]* y[2] + x[3]* y[2] + x[1]* y[3] - x[2]* y[3]));        b = -(-x[1]*x[1]* x[2] + x[1]*x[2]*x[2] + x[1]*x[1] *x[3] - x[2]*x[2]* x[3] - x[1]* x[3]*x[3] + x[2]*  x[3]*x[3] - x[2] *y[1]*y[1] + x[3] *y[1]*y[1] + x[1] *y[2]*y[2] - x[3] *y[2]*y[2] - x[1]* y[3]*y[3] + x[2]* y[3]*y[3])/(2 *(x[2] *y[1] - x[3]* y[1] - x[1] *y[2] + x[3] *y[2] + x[1] *y[3] - x[2]* y[3]));        r = sqrt((x[1]-a)*(x[1]-a) + (y[1]-b)*(y[1]-b));        //printf("%lf %lf %lf\n",a,b,r);        double t = sqrt((x1-a)*(x1-a) + (y1-b)*(y1-b));        if(t>r)            printf("Accepted\n");        else            printf("Rejected\n");    }    return 0;}