-
-
Save paranoidxc/3f6e48041501f3b4397c to your computer and use it in GitHub Desktop.
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
/* | |
(x+1)的N次方的多项式 | |
*/ | |
#include <stdio.h> | |
#include <stdlib.h> | |
#include "POLY.h" | |
main(int argc, char *argv[]) | |
{ int N = atoi(argv[1]); float p = atof(argv[2]); | |
Poly t, x; int i, j; | |
printf("Binomial coefficients\n"); | |
t = POLYadd(POLYterm(1, 1), POLYterm(1, 0)); | |
for (i = 0, x = t; i < N; i++) | |
{ x = POLYmult(t, x); showPOLY(x); } | |
printf("%f\n", POLYeval(x, p)); | |
} | |
----- | |
typedef struct poly *Poly; | |
void showPOLY(Poly); | |
Poly POLYterm(int, int); | |
Poly POLYadd(Poly, Poly); | |
Poly POLYmult(Poly, Poly); | |
float POLYeval(Poly, float); | |
----- | |
#include <stdlib.h> | |
#include "POLY.h" | |
struct poly { int N; int *a; }; | |
Poly POLYterm(int coeff, int exp) | |
{ int i; Poly t = malloc(sizeof *t); | |
t->a = malloc((exp+1)*sizeof(int)); | |
t->N = exp+1; t->a[exp] = coeff; | |
for (i = 0; i < exp; i++) t->a[i] = 0; | |
return t; | |
} | |
Poly POLYadd(Poly p, Poly q) | |
{ int i; Poly t; | |
if (p->N < q->N) { t = p; p = q; q = t; } | |
for (i = 0; i < q->N; i++) p->a[i] += q->a[i]; | |
return p; | |
} | |
Poly POLYmult(Poly p, Poly q) | |
{ int i, j; | |
Poly t = POLYterm(0, (p->N-1)+(q->N-1)); | |
for (i = 0; i < p->N; i++) | |
for (j = 0; j < q->N; j++) | |
t->a[i+j] += p->a[i]*q->a[j]; | |
return t; | |
} | |
float POLYeval(Poly p, float x) | |
{ int i; double t = 0.0; | |
for (i = p->N-1; i >= 0; i--) | |
t = t*x + p->a[i]; | |
return t; | |
} |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
缺少 showPOLY() 的实现,话说也没弄明白这个多项式是怎么计算的。
说是计算 (x+1) 的 N 次方,但是怎么和手算出来的不一样 - -!