Simpson公式的应用（HDU 1724/ HDU 1071）

辛普森积分法 - 维基百科，自由的百科全书

Simpson's rule - Wikipedia, the free encyclopedia

　　利用这个公式，用二分的方法来计算积分。

1071 ( The area )

 #include <iostream>
 #include <algorithm>
 #include <cstdio>
 #include <cstring>
 #include <cmath>

 using namespace std;

 const double EPS = 1e-;
 double A, B, C, P, Q;

 template<class T> T sqr(T x) { return x * x;}
 inline double cal(double x) { return A * sqr(x) + (B - P) * x + C - Q;}
 inline double sps(double l, double r) { return (cal(l) + cal(r) +  * cal((l + r) / )) /  * (r - l);}

 double work(double l, double r) {
     //cout << l << ' ' << r << endl;
     double ans = sps(l, r), m = (l + r) / ;
     if (fabs(ans - sps(l, m) - sps(m, r)) < EPS) return ans;
     else return work(l, m) + work(m, r);
 }

 int main() {
     int T;
     double l, r;
     double x[], y[];
     cin >> T;
     while (T--) {
         for (int i = ; i < ; i++) cin >> x[i] >> y[i];
         double p[], q[], d[];
         for (int i = ; i < ; i++) p[i] = sqr(x[i]) - sqr(x[i + ]), q[i] = x[i] - x[i + ], d[i] = y[i] - y[i + ];
         A = (q[] * d[] - q[] * d[]) / (p[] * q[] - p[] * q[]);
         B = (p[] * d[] - p[] * d[]) / (p[] * q[] - p[] * q[]);
         C = y[] - B * x[] - A * sqr(x[]);
         //cout << A << ' ' << B << ' ' << C << endl;
         P = (y[] - y[]) / (x[] - x[]);
         Q = y[] - P * x[];
         //cout << P << ' ' << Q << endl;
         printf("%.2f\n", work(x[], x[]));
     }
     return ;
 }

1724 ( Ellipse )

 #include <iostream>
 #include <algorithm>
 #include <cstdio>
 #include <cstring>
 #include <cmath>

 using namespace std;

 const double EPS = 1e-;
 double A, B;

 template<class T> T sqr(T x) { return x * x;}
 inline double cal(double x) { return  * B * sqrt( - sqr(x) / sqr(A));}
 inline double sps(double l, double r) { return (cal(l) + cal(r) +  * cal((l + r) / )) /  * (r - l);}

 double work(double l, double r) {
     //cout << l << ' ' << r << endl;
     double ans = sps(l, r), m = (l + r) / ;
     if (fabs(ans - sps(l, m) - sps(m, r)) < EPS) return ans;
     else return work(l, m) + work(m, r);
 }

 int main() {
     int T;
     double l, r;
     cin >> T;
     while (T-- && cin >> A >> B >> l >> r) printf("%.3f\n", work(l, r));
     return ;
 }

　　之后还有题会继续更新。

UPD：

　　就是因为见过这题，所以才学这个公式的。1y~

ACM-ICPC Live Archive

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 #include <cmath>

 using namespace std;

 double coe[][];
 const double EPS = 1e-;

 int k;
 double cal(double x, double *c) {
     double ret = c[];
     for (int i = ; i <= k; i++) ret *= x, ret += c[i];
     return ret;
 }

 inline double cal(double x, double *p, double *q) { return cal(x, p) / cal(x, q);}
 inline double cal(double x, double y, double *p, double *q) { return max(cal(x, p, q) - y, 0.0);}
 inline double simpson(double y, double l, double r, double *p, double *q) { return (cal(l, y, p, q) + cal(r, y, p, q) +  * cal((l + r) / , y, p, q)) * (r - l) / ;}

 inline double getpart(double y, double l, double r, double *p, double *q) {
     double sum = simpson(y, l, r, p, q);
     //cout << l << ' ' << r << ' ' << sum << endl;
     if (fabs(sum - simpson(y, l, (l + r) / , p, q) - simpson(y, (l + r) / , r, p, q)) < EPS) return sum;
     return getpart(y, l, (l + r) / , p, q) + getpart(y, (l + r) / , r, p, q);
 }

 inline double getarea(double y, double l, double r, double *p, double *q) {
     double ret = , d = (r - l) / ;
     for (int i = ; i < ; i++) {
         ret += getpart(y, l + d * i, l + d * (i + ), p, q);
     }
     return ret;
 }

 double dc2(double l, double r, double a, double w) {
     double m;
     while (r - l > EPS) {
         m = (l + r) / 2.0;
         //cout << m << ' ' << getarea(m, 0, w, coe[0], coe[1]) - getarea(m, 0, w, coe[2], coe[3]) << endl;
         if (getarea(m, , w, coe[], coe[]) - getarea(m, , w, coe[], coe[]) > a) l = m;
         else r = m;
     }
     return l;
 }

 int main() {
     //freopen("in", "r", stdin);
     //freopen("out", "w", stdout);
     double w, d, a;
     while (cin >> w >> d >> a >> k) {
         for (int i = ; i < ; i++) for (int j = ; j <= k; j++) cin >> coe[i][j];
         for (int i = ; i < ; i++) reverse(coe[i], coe[i] + k + );
         //cout << getarea(-5.51389, 0, w, coe[0], coe[1]) - getarea(-5.51389, 0, w, coe[2], coe[3]) << endl;
         //cout << cal(3, coe[0], coe[1]) << endl;
         printf("%.5f\n", -dc2(-d, , a, w));
     }
     return ;
 }

——written by Lyon