http://acm.hdu.edu.cn/showproblem.php?pid=5060
官方题解http://bestcoder.hdu.edu.cn/给复杂了
实际上用圆柱体与球体体积差的积分公式(pi*r*r - pi*x*x)即可轻松解决五种情况
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define RD(x) scanf("%d",&x)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define clr0(x) memset(x,0,sizeof(x))
typedef long long LL;
const double pi = acos ( -1.0 ) ;
int R , HR , HZ ;
double f2 ( double x1 , double x2 ) {
return R * R * pi * ( x2 - x1 ) - 1.0 / 3.0 * pi * ( x2 * x2 * x2 - x1 * x1 * x1 ) ;
}
int solve () {
double V1 = pi * R * R * R * 4.0 / 3.0 ;
double V2 = 2.0 * HR * HR * pi * HZ ;
if ( HR * HR + HZ * HZ <= R * R ) {
printf ( "%.6f\n" , V2 / V1 ) ;
} else if ( HR >= R && HZ >= R ) {
printf ( "%.6f\n" , V1 / V2 ) ;
} else {
double V;
if ( HR <= R && HZ <= R ) {
double y1 = HZ ;
double y2 = sqrt ( R * R - HR * HR ) ;
V = 2.0 * ( HR * HR * y2 * pi + f2 ( y2 , R ) - f2 ( y1 , R ) ) ;
} else if ( HR > R && HZ <= R ) {
double y1 = HZ ;
double y2 = 0 ;
V = 2.0 * ( f2 ( y2 , R ) - f2 ( y1 , R ) ) ;
} else if ( HR <= R && HZ >= R ) {
double y = sqrt ( R * R - HR * HR ) ;
V = 2.0 * ( HR * HR * y * pi + f2 ( y , R ) ) ;
}
printf ( "%.6f\n" , V / ( V1 + V2 - V ) ) ;
}
}
int main () {
while ( ~RD3(R , HR , HZ ) )
solve () ;
return 0 ;
}