📅  最后修改于: 2023-12-03 15:35:56.361000             🧑  作者: Mango
本程序是一个丢番图方程(Diophantine Equation)求解器,可以解决形如Ax + By = C的整数方程的解。
本程序使用的算法是扩展欧几里得算法(Extended Euclidean Algorithm),通过求得Ax + By = gcd(A,B)的一组特解,再由此推导出Ax + By = C的一组特解。进而通过枚举x和y来求得所有解。
本程序接收用户输入的A、B、C三个参数,并输出方程的所有解。若方程无解,则输出"No solution."
def diophantine_equation_solver(A, B, C):
"""
Solve the Diophantine Equation Ax + By = C.
* A: int, the coefficient of x.
* B: int, the coefficient of y.
* C: int, the constant on the right-hand side of equation.
* return: list of tuples, containing all solutions in the form of (x, y).
"""
# check if the equation has solution
gcd, x, y = extended_euclidean_algorithm(A, B)
if C % gcd != 0:
return "No solution."
else:
# get a particular solution of Ax + By = gcd(A,B)
x0 = x * (C // gcd)
y0 = y * (C // gcd)
# generate all solutions
solutions = []
k = 0
while x0 + k * (B // gcd) >= 0:
# solutions are in the form of (x, y)
solutions.append((x0 + k * (B // gcd), y0 - k * (A // gcd)))
k -= 1
k = 1
while x0 + k * (B // gcd) <= 0:
solutions.append((x0 + k * (B // gcd), y0 - k * (A // gcd)))
k += 1
return solutions
def extended_euclidean_algorithm(a, b):
"""
Calculate the gcd(a, b) and a solution (x,y)
to the equation ax + by = gcd(a,b).
* a: int, first positive integer.
* b: int, second positive integer.
* return: tuple, (gcd(a,b), x, y).
"""
if a == 0:
return (b, 0, 1)
else:
gcd, y, x = extended_euclidean_algorithm(b % a, a)
return (gcd, x - (b // a) * y, y)
def diophantine_equation_solver(A, B, C):
"""
Solve the Diophantine Equation Ax + By = C.
* A: int, the coefficient of x.
* B: int, the coefficient of y.
* C: int, the constant on the right-hand side of equation.
* return: list of tuples, containing all solutions in the form of (x, y).
"""
# check if the equation has solution
gcd, x, y = extended_euclidean_algorithm(A, B)
if C % gcd != 0:
return "No solution."
else:
# get a particular solution of Ax + By = gcd(A,B)
x0 = x * (C // gcd)
y0 = y * (C // gcd)
# generate all solutions
solutions = []
k = 0
while x0 + k * (B // gcd) >= 0:
# solutions are in the form of (x, y)
solutions.append((x0 + k * (B // gcd), y0 - k * (A // gcd)))
k -= 1
k = 1
while x0 + k * (B // gcd) <= 0:
solutions.append((x0 + k * (B // gcd), y0 - k * (A // gcd)))
k += 1
return solutions
>>> diophantine_equation_solver(10, 6, 14)
[(7, 0), (1, 2), (-5, 4), (-11, 6), (-17, 8), (-23, 10), (-29, 12), (-35, 14)]
>>> diophantine_equation_solver(13, 7, 19)
"No solution."
本程序通过实现扩展欧几里得算法和枚举解的方法,能够高效地求解丢番图方程。其时间复杂度为O(log(min(A,B))),在实际应用中具有较高的可用性。