math专辑

1.hdoj 3310 Volume of a cylinder

求两个圆柱相交部分的体积。

第一象限积分公式：sqrt(R * R - (r - x) * (r - x)) * sqrt(r * r - (r - x) * (r - x));   从0积到r 。最后结果乘8。

可是不知道为什么我用龙贝格积分却错了，用分割小块就AC了，忘路过大牛指点。

分割小块：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

const int cntStep = 1000000;

double solve(double R, double r) {
    double step, ans = 0;
    double w, h, pos;
    int i;
    step = r / (cntStep * 1.0);
    for(i=0; i<cntStep; ++i) {
        pos = i * step + step/2;
        h = sqrt(R * R - (r - pos) * (r - pos));
        w = sqrt(r * r - (r - pos) * (r - pos));
        ans += w * h;
    }
    return ans * step * 8;
}


int main() {
	//    freopen("c:/aaa.txt", "r", stdin);
    int T;
    double temp, R, r;
    scanf("%d", &T);
    while(T --) {
        scanf("%lf %lf", &R, &r);
        if(R < r) {
            temp = R;
            R = r;
            r = temp;
        }
        
        printf("%.2lf\n", solve(R, r));
    }
    return 0;
}

  龙贝格法：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;


#define ep    0.000001     //精度限制
const int M = 100;
double ans[M][M];
double R, r;


double f(double x) {
    return 4 * sqrt(R * R - (r - x) * (r - x)) * sqrt(r * r - (r - x) * (r - x));
}


double Romberg(double a,double b) { //从a积到b
    double h,temp;
    double v;
    int i,j,k,n;
    bool flag=1;

    h = b - a;
    ans[0][0] = h/2 * ( f(a) + f(b) );
    for(i=1;i<20;++i){
        for(j=0;j<=i;++j){
            if(j==0) {
                ans[i][0] = ans[i-1][0]/2;
                n = (int)pow(2.0,i);

                for(k=1,temp=0;k<=n/2;++k){
                    v = h/n*(2*k-1) + a;
                    temp+=f(v);
                }
                
                temp = temp*h/n;
                ans[i][0]+=temp;
            }
            else{
                n=(int)pow(4.0,j);
                ans[i][j] = (ans[i][j-1]*n - ans[i-1][j-1])/(n-1);
            }
        }
        temp = fabs(ans[i][i] - ans[i-1][i-1]) ;
        if(temp <= ep){
            return ans[i][i];
        }
    }
    return ans[19][19];
}


int main() {
//    freopen("c:/aaa.txt", "r", stdin);
    int T;
    double temp;
    scanf("%d", &T);
    while(T --) {
        scanf("%lf %lf", &R, &r);
        if(R < r) {
            temp = R;
            R = r;
            r = temp;
        }
        printf("%.2lf\n", 2 * Romberg(0, r));
    }
    return 0;
}