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  1. #include <iostream>
  2. #include <math.h>
  3.  
  4. using namespace std;
  5.  
  6. #define PI 4*atan(1)
  7.  
  8.  
  9. double scal(double * x, double * y, int N, double h) {
  10. 	double t = 0.0;
  11. 	for (int i = 1; i < N; i++) {
  12. 		t+= x[i] * y[i] * h;
  13. 	}
  14. 	t += (h/2) * (x[0]*y[0] + x[N]*y[N]);
  15. 	return t;
  16. }
  17.  
  18. void f2c(double * f, int N, double h) {
  19. 	double * c = (double *) malloc(sizeof(double) * (N - 1));
  20. 	double * phi = (double *) malloc(sizeof(double) * (N + 1));
  21. 	for (int i = 0; i < N - 1; i++) {
  22. 		for (int j = 0; j < N + 1; j++) {
  23. 			double x = (j - 0.5) * h;
  24. 			phi[j] = sqrt(2) * sin(PI * x * (i + 1));
  25. 		}
  26. 		double l = 4.0 / (h*h) * sin(PI * h * (i + 1)) * sin(PI * h * (i + 1));
  27. 		c[i] = scal(f,phi,N,h)/l;
  28. 	}
  29. 	for (int i = 0; i < N + 1; i++) {
  30. 		double t = 0.0;
  31. 		for (int j = 0; j < N - 1; j++) {
  32. 			t += c[j] * sqrt(2) * sin(PI * (j + 1) * (i - 0.5) * h);
  33. 		}
  34. 		cout << t << " " << f[i] << endl;
  35. 	}
  36. }
  37.  
  38.  
  39. int main() {
  40. 	int N;
  41. 	cin >> N;
  42. 	//double h = 1.0 / (N - 1);
  43. 	double h = 1.0 / (N);
  44. 	double * f = (double *) malloc(sizeof(double) * (N - 1));
  45. 	double * phi = (double *) malloc(sizeof(double) * (N + 1));
  46. 	double * psi = (double *) malloc(sizeof(double) * (N + 1));
  47. 	for (int i = 0; i < N - 1; i++) {
  48. 		for (int j = 0; j < N + 1; j++) {
  49. 			phi[j] = sqrt(2) * sin(PI * (j-0.5) * h);
  50. 			psi[j] = sqrt(2) * sin(PI * (j-0.5) *(i + 1) * h);
  51. 		}
  52. 		cout << scal(phi,psi,N,h) << endl;
  53. 	}
  54. 	for (int i = 0; i < N - 1; i++) {
  55. 		f[i] = sin(PI * (i - 0.5) * h);
  56. 	}
  57. 	//f2c(f,N,h);
  58. 	return 0;
  59. }
Success #stdin #stdout 0s 4388KB
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stdin
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8
stdout
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1
0.0186645
-1.07553e-16
0.0344874
9.71445e-17
0.04506
8.88178e-16
https://ideone.com/ildpq3
language:
C++14 (gcc 8.3)
created:
4 years ago
visibility:
public

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