44 lines
1 KiB
C
44 lines
1 KiB
C
#include <stdio.h>
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#include <time.h>
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#include <math.h>
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#include <float.h>
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#include "init_f.h"
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#include "solve.h"
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/* ./a05.out a b n k */
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int main (int argc, char *argv[])
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{
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double t, integral, a, b;
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int k, n, calls, task = 5;
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double (*f_lst[]) (double) = {f0, f1, f2, f3, f4, f5, f6};
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int len_f = sizeof(f_lst) / sizeof(f_lst[0]);
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if (
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!((argc == 5) &&
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sscanf(argv[1], "%lf", &a) == 1 &&
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sscanf(argv[2], "%lf", &b) == 1 &&
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((sscanf(argv[3], "%d", &n) == 1) && (n > 0)) &&
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((sscanf(argv[4], "%d", &k) == 1) && ((0 <= k) && (k < len_f))))
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) {
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fprintf(stderr, "Usage: %s a b n k\n", argv[0]);
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return -1;
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}
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t = clock();
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integral = t5_solve(f_lst[k], a, b, n);
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t = (clock() - t) / CLOCKS_PER_SEC;
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calls = get_call_count();
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if (fabs(integral - DBL_MAX) < DBL_EPSILON) {
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fprintf (stdout, "%s : Task = %d Method is not applicable Count = %d T = %.2f\n", argv[0], task, calls, t);
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return -2;
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} else {
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fprintf (stdout, "%s : Task = %d Res = %e Count = %d T = %.2f\n", argv[0], task, integral, calls, t);
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return 0;
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}
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}
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