題目意思很簡(jiǎn)單，求拋物線和直線圍成的區(qū)域的面積。
積分大牛都是0msAC的……不怎么會(huì)積分，于是直接龍貝格定積分求值。
以下是我的代碼：
/*
* Author: lee1r
* Created Time: 2011/8/23 18:19:49
* File Name: hdu1071.cpp
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#define L(x) ((x)<<1)
#define R(x) ((x)<<1|1)
#define Half(x) ((x)>>1)
#define Lowbit(x) ((x)&(-(x)))
using namespace std;
const int kInf(0x7f7f7f7f);
const double kEps(1e-8);
typedef unsigned int uint;
typedef long long int64;
typedef unsigned long long uint64;
bool scanf(int &num)
{
char in;
while((in=getchar())!=EOF && (in>'9' || in<'0'));
if(in==EOF) return false;
num=in-'0';
while(in=getchar(),in>='0' && in<='9') num*=10,num+=in-'0';
return true;
}
double A,B,C;
double f(double x)
{
return (A*x*x+B*x+C);
}
double Romberg(double a,double b,double eps)
{
#define MAX_N 1000
int min,tmp2;
double d,tmp4,R[2][MAX_N];
for(int i=0;i<MAX_N;i++)
R[0][i]=R[1][i]=.0;
d=b-a;
min=(int)(log(d*10.0)/log(2.0));
R[0][0]=0.5*d*(f(a)+f(b));
tmp2=1;
for(int i=2;i<MAX_N;i++)
{
R[1][0]=.0;
for(int j=1;j<=tmp2;j++)
R[1][0]+=f(a+d*((double)j-0.5));
R[1][0]=(R[0][0]+d*R[1][0])*0.5;
tmp4=4.0;
for(int j=1;j<i;j++)
{
R[1][j]=R[1][j-1]+(R[1][j-1]-R[0][j-1])/(tmp4-1.0);
tmp4*=4.0;
}
if((fabs(R[1][i-1]-R[0][i-2])<eps) && (i>min))
return R[1][i-1];
d*=0.5;
tmp2*=2;
for(int j=0;j<i;j++)
R[0][j]=R[1][j];
}
return R[1][MAX_N-1];
}
int main()
{
#ifndef ONLINE_JUDGE
//freopen("data.in","r",stdin);
#endif
int T;
scanf(T);
while(T--)
{
double x0,y0,x1,y1,x2,y2;
scanf("%lf%lf%lf%lf%lf%lf",&x0,&y0,&x1,&y1,&x2,&y2);
B=(y2-y1-(x2*x2-x1*x1)*(y1-y0)/(x1*x1-x0*x0))/(x2-x1-(x2*x2-x1*x1)/(x1+x0));
A=(y1-y0-(x1-x0)*B)/(x1*x1-x0*x0);
C=y0-A*x0*x0-B*x0;
B-=(y2-y1)/(x2-x1);
C+=x1*(y2-y1)/(x2-x1)-y1;
printf("%.2f\n",Romberg(x1,x2,kEps));
}
return 0;
}