2nd_Sem_Bogachev/2025.05.09/10Ex/main.c
2025-05-14 20:12:23 +03:00

42 lines
1,011 B
C

#include <stdio.h>
#include <time.h>
#include <math.h>
#include <float.h>
#include "init_f.h"
#include "solve.h"
/* ./a10.out a eps k */
int main (int argc, char *argv[])
{
double t, integral, a, b, eps;
int k, calls, task = 10;
double (*f_lst[]) (double) = {f0, f1, f2, f3, f4, f5, f6};
int len_f = sizeof(f_lst) / sizeof(f_lst[0]);
if (
!((argc == 4) &&
sscanf(argv[1], "%lf", &a) == 1 &&
((sscanf(argv[2], "%lf", &eps) == 1) && (eps > 0)) &&
((sscanf(argv[3], "%d", &k) == 1) && ((0 <= k) && (k < len_f))))
) {
fprintf(stderr, "Usage: %s a eps k\n", argv[0]);
return -1;
}
t = clock();
b = t10_solve(f_lst[k], a, eps, &integral);
t = (clock() - t) / CLOCKS_PER_SEC;
calls = get_call_count();
if (b < 0) {
fprintf (stdout, "%s : Task = %d Method is not applicable Count = %d T = %.2f\n", argv[0], task, calls, t);
return -2;
} else {
fprintf (stdout, "%s : Task = %d Res = %e B = %e Count = %d T = %.2f\n", argv[0], task, integral, b, calls, t);
return 0;
}
}