利用这个公式，用二分的方法来计算积分。
1 #include2 #include 3 #include 4 #include 5 #include 6 7 using namespace std; 8 9 const double EPS = 1e-8;10 double A, B, C, P, Q;11 12 template T sqr(T x) { return x * x;}13 inline double cal(double x) { return A * sqr(x) + (B - P) * x + C - Q;}14 inline double sps(double l, double r) { return (cal(l) + cal(r) + 4 * cal((l + r) / 2)) / 6 * (r - l);}15 16 double work(double l, double r) {17 //cout << l << ' ' << r << endl;18 double ans = sps(l, r), m = (l + r) / 2;19 if (fabs(ans - sps(l, m) - sps(m, r)) < EPS) return ans;20 else return work(l, m) + work(m, r);21 }22 23 24 int main() {25 int T;26 double l, r;27 double x[3], y[3];28 cin >> T;29 while (T--) {30 for (int i = 0; i < 3; i++) cin >> x[i] >> y[i];31 double p[2], q[2], d[2];32 for (int i = 0; i < 2; i++) p[i] = sqr(x[i]) - sqr(x[i + 1]), q[i] = x[i] - x[i + 1], d[i] = y[i] - y[i + 1];33 A = (q[1] * d[0] - q[0] * d[1]) / (p[0] * q[1] - p[1] * q[0]);34 B = (p[1] * d[0] - p[0] * d[1]) / (p[1] * q[0] - p[0] * q[1]);35 C = y[0] - B * x[0] - A * sqr(x[0]);36 //cout << A << ' ' << B << ' ' << C << endl;37 P = (y[1] - y[2]) / (x[1] - x[2]);38 Q = y[1] - P * x[1];39 //cout << P << ' ' << Q << endl;40 printf("%.2f\n", work(x[1], x[2]));41 }42 return 0;43 }
1 #include2 #include 3 #include 4 #include 5 #include 6 7 using namespace std; 8 9 const double EPS = 1e-8;10 double A, B;11 12 template T sqr(T x) { return x * x;}13 inline double cal(double x) { return 2 * B * sqrt(1 - sqr(x) / sqr(A));}14 inline double sps(double l, double r) { return (cal(l) + cal(r) + 4 * cal((l + r) / 2)) / 6 * (r - l);}15 16 double work(double l, double r) {17 //cout << l << ' ' << r << endl;18 double ans = sps(l, r), m = (l + r) / 2;19 if (fabs(ans - sps(l, m) - sps(m, r)) < EPS) return ans;20 else return work(l, m) + work(m, r);21 }22 23 int main() {24 int T;25 double l, r;26 cin >> T;27 while (T-- && cin >> A >> B >> l >> r) printf("%.3f\n", work(l, r));28 return 0;29 }
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~
1 #include2 #include 3 #include 4 #include 5 #include 6 7 using namespace std; 8 9 double coe[4][11];10 const double EPS = 1e-8;11 12 int k;13 double cal(double x, double *c) {14 double ret = c[0];15 for (int i = 1; i <= k; i++) ret *= x, ret += c[i];16 return ret;17 }18 19 inline double cal(double x, double *p, double *q) { return cal(x, p) / cal(x, q);}20 inline double cal(double x, double y, double *p, double *q) { return max(cal(x, p, q) - y, 0.0);}21 inline double simpson(double y, double l, double r, double *p, double *q) { return (cal(l, y, p, q) + cal(r, y, p, q) + 4 * cal((l + r) / 2, y, p, q)) * (r - l) / 6;}22 23 inline double getpart(double y, double l, double r, double *p, double *q) {24 double sum = simpson(y, l, r, p, q);25 //cout << l << ' ' << r << ' ' << sum << endl;26 if (fabs(sum - simpson(y, l, (l + r) / 2, p, q) - simpson(y, (l + r) / 2, r, p, q)) < EPS) return sum;27 return getpart(y, l, (l + r) / 2, p, q) + getpart(y, (l + r) / 2, r, p, q);28 }29 30 inline double getarea(double y, double l, double r, double *p, double *q) {31 double ret = 0, d = (r - l) / 100;32 for (int i = 0; i < 100; i++) {33 ret += getpart(y, l + d * i, l + d * (i + 1), p, q);34 }35 return ret;36 }37 38 double dc2(double l, double r, double a, double w) {39 double m;40 while (r - l > EPS) {41 m = (l + r) / 2.0;42 //cout << m << ' ' << getarea(m, 0, w, coe[0], coe[1]) - getarea(m, 0, w, coe[2], coe[3]) << endl;43 if (getarea(m, 0, w, coe[0], coe[1]) - getarea(m, 0, w, coe[2], coe[3]) > a) l = m;44 else r = m;45 }46 return l;47 }48 49 int main() {50 //freopen("in", "r", stdin);51 //freopen("out", "w", stdout);52 double w, d, a;53 while (cin >> w >> d >> a >> k) {54 for (int i = 0; i < 4; i++) for (int j = 0; j <= k; j++) cin >> coe[i][j];55 for (int i = 0; i < 4; i++) reverse(coe[i], coe[i] + k + 1);56 //cout << getarea(-5.51389, 0, w, coe[0], coe[1]) - getarea(-5.51389, 0, w, coe[2], coe[3]) << endl;57 //cout << cal(3, coe[0], coe[1]) << endl;58 printf("%.5f\n", -dc2(-d, 0, a, w));59 }60 return 0;61 }
——written by Lyon