1 solutions

  • 0
    @ 2025-11-5 20:08:53

    C :

    #include<stdio.h>
    
    int main()
    {
    	int t;
    	float xp1,xp2,xp3,yp1,yp2,yp3;
    	double a,k,b,s;
    	scanf("%d",&t);
    	while(t--)
    	{
    		scanf("%f%f%f%f%f%f",&xp1,&yp1,&xp2,&yp2,&xp3,&yp3);
    		a=(yp2-yp1)/(xp2-xp1)/(xp2-xp1);
    		k=(yp2-yp3)/(xp2-xp3);
    		b=yp2-(yp2-yp3)/(xp2-xp3)*xp2;
    		s=a/3*xp3*xp3*xp3+(-2*a*xp1)/2*xp3*xp3+(a*xp1*xp1+yp1)*xp3-k/2*xp3*xp3-b*xp3-(a/3*xp2*xp2*xp2+(-2*a*xp1)/2*xp2*xp2+(a*xp1*xp1+yp1)*xp2-k/2*xp2*xp2-b*xp2);
    		printf("%.2lf\n",s);
    	}
    	return 0;
    }
    

    C++ :

    #include<stdio.h>
    
    int main()
    {
    	int t;
    	float xp1,xp2,xp3,yp1,yp2,yp3;
    	double a,k,b,s;
    	scanf("%d",&t);
    	while(t--)
    	{
    		scanf("%f%f%f%f%f%f",&xp1,&yp1,&xp2,&yp2,&xp3,&yp3);
    		a=(yp2-yp1)/(xp2-xp1)/(xp2-xp1);
    		k=(yp2-yp3)/(xp2-xp3);
    		b=yp2-(yp2-yp3)/(xp2-xp3)*xp2;
    		s=a/3*xp3*xp3*xp3+(-2*a*xp1)/2*xp3*xp3+(a*xp1*xp1+yp1)*xp3-k/2*xp3*xp3-b*xp3-(a/3*xp2*xp2*xp2+(-2*a*xp1)/2*xp2*xp2+(a*xp1*xp1+yp1)*xp2-k/2*xp2*xp2-b*xp2);
    		printf("%.2lf\n",s);
    	}
    	return 0;
    }
    
    • 1

    Information

    ID
    20059
    Time
    1000ms
    Memory
    32MiB
    Difficulty
    (None)
    Tags
    # Submissions
    0
    Accepted
    0
    Uploaded By