hdu 5060 五种情况求圆柱体与球体交

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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 <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <queue>#include <set>#include <iostream>#include <algorithm>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 ;}

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