1、http://acm.hdu.edu.cn/showproblem.php?pid=1071
2、题目:
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6511 Accepted Submission(s): 4566
2 5.000000 5.000000 0.000000 0.000000 10.000000 0.000000 10.000000 10.000000 1.000000 1.000000 14.000000 8.222222
33.33 40.69
3、Ac代码
#include<stdio.h> int main() { int t; double x1,y1,x2,y2,x3,y3,a,b,c,K,B; scanf("%d",&t); while(t--) { scanf("%lf%lf",&x1,&y1); scanf("%lf%lf",&x2,&y2); scanf("%lf%lf",&x3,&y3); a=(y2-y1)/((x2-x1)*(x2-x1)); b=(y2-y3-(a*(x2*x2-x3*x3)))/(x2-x3); c=y2-a*x2*x2-b*x2; K=(y2-y3)/(x2-x3); B=y2-K*x2; //printf("%lf %lf %lf\n",a,b,c); double ans1=((a/3)*x3*x3*x3+(b/2)*x3*x3+c*x3)-((a/3)*x2*x2*x2+(b/2)*x2*x2+c*x2); double ans2=((K/2)*x3*x3+B*x3)-((K/2)*x2*x2+B*x2); //printf("**%lf %lf\n",ans1,ans2); printf("%.2lf\n",ans1-ans2); } return 0; }