1 solutions
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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; }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