HDU 1071 The area

HDU 1071 The area

题目意思很简单，求抛物线和直线围成的区域的面积。
积分大牛都是0msAC的……不怎么会积分，于是直接龙贝格定积分求值。
以下是我的代码：

/*
 * 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 ;
}