Created
June 16, 2025 18:21
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Computes GCD and solves the linear Diophantine equation ax + by = c with Bézout coefficients.
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#include <assert.h> | |
#include <stdio.h> | |
#include <stdlib.h> | |
int iabs(int a) { return (a < 0) ? -a : a; } | |
int eu_mod(int a, int b) { | |
int r; | |
assert(b != 0); | |
r = a % b; | |
if (r < 0) | |
r += iabs(b); | |
return r; | |
} | |
int eu_div(int a, int b) { | |
int q; | |
assert(b != 0); | |
q = a / b; | |
if (a % b < 0 && b < 0) | |
q += 1; | |
else if (a % b < 0) | |
q -= 1; | |
return q; | |
} | |
int gcd(int a, int b) { | |
int q; | |
q = eu_mod(a, b); | |
while(q != 0) { | |
a = b; | |
b = q; | |
q = eu_mod(a, q); | |
} | |
return b; | |
} | |
int gcd_ext(int a, int b, int* x, int* y) { | |
int r2; | |
int x0 = 1, x1 = 0, y0 = 0, y1 = 1; | |
r2 = eu_mod(a, b); | |
*x = x0 - x1 * eu_div(a, b); | |
*y = y0 - y1 * eu_div(a, b); | |
if(r2 == 0) { | |
*x = 0; | |
*y = 1; | |
} | |
while(r2 != 0) { | |
a = b; | |
b = r2; | |
r2 = eu_mod(a, r2); | |
if(r2 != 0) { | |
x0 = x1; | |
x1 = *x; | |
*x = x0 - x1 * eu_div(a, b); | |
y0 = y1; | |
y1 = *y; | |
*y = y0 - y1 * eu_div(a, b); | |
} | |
} | |
return b; | |
} | |
void lde(int a, int b, int c, int* x, int* y) { | |
int d; | |
d = gcd_ext(a, b, &*x, &*y); | |
*x = (c/d) * (*x); | |
*y = (c/d) * (*y); | |
} | |
int main(void) { | |
int a = 0, b = 0, c = 0, x, y, d, u, v; | |
int res; | |
printf("Enter values of a, b, and c in equation ax + by = c: \n"); | |
res = scanf("%i %i %i", &a, &b, &c); | |
assert(res == 3); | |
d = gcd_ext(a, b, &x, &y); | |
printf("GCD(%i, %i) = %i\n", a, b, d); | |
printf("Bézout's coefficients for equation %ix + %iy = %i: x = %i, y = %i\n", a, b, d, x, y); | |
if(c % gcd(a, b) != 0) { | |
printf("Linear diophantine equation %iu + %iv = %i is not solvable\n", a, b, c); | |
return 0; | |
} | |
lde(a, b, c, &u, &v); | |
printf("Coefficients for linear diophantine equation %iu + %iv = %i: u = %i, v = %i\n", a, b, c, u, v); | |
return 0; | |
} |
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