Question

我将问题简化为以下内容：

Xn + 1 = Xn + Yn

Yn + 1 = Yn + Zn

Zn + 1 = Zn + Xn

我知道X0，Y0，Z0的值等于1。我想告诉python找到X1，Y1，Z1的值，然后找到X2，Y2，Z2，......等。任何人都可以帮助我吗？我想我必须使用嵌套循环，但我不确定如何去做。谢谢！

Answer 1

我们在谈论的事情很简单：

x, y, z = 1, 1, 1

for i in range(10):
    print("X{i} = {x}, Y{i} = {y}, Z{i} = {z}".format(**locals()))

    x, y, z = x + y, y + z, z + x

由于输出无趣，所以看起来并不正确：

X0 = 1, Y0 = 1, Z0 = 1
X1 = 2, Y1 = 2, Z1 = 2
X2 = 4, Y2 = 4, Z2 = 4
X3 = 8, Y3 = 8, Z3 = 8
X4 = 16, Y4 = 16, Z4 = 16
X5 = 32, Y5 = 32, Z5 = 32
X6 = 64, Y6 = 64, Z6 = 64
X7 = 128, Y7 = 128, Z7 = 128
X8 = 256, Y8 = 256, Z8 = 256
X9 = 512, Y9 = 512, Z9 = 512

Answer 2

以下是实现它的示例函数：

def solve_equation(n):
    X = {0: 1}
    Y = {0: 1}
    Z = {0: 1}
    for i in range(n):
        print 'For n: ', i+1
        X[i+1] = X[i] + Y[i]
        Y[i+1] = Y[i] + Z[i]
        Z[i+1] = Z[i] + X[i]
        print 'X = ', X[i+1], ' Y = ', Y[i+1], ' Z = ', Z[i+1]

示例运行：

>>> solve_equation(3)
For n:  1
X =  2  Y =  2  Z =  2
For n:  2
X =  4  Y =  4  Z =  4
For n:  3
X =  8  Y =  8  Z =  8

Answer 3

您可以使用递归函数。要获得更快更好的代码，您可以考虑使用 yield

def Cvalue(c,xyz , n):
     if n == 0 :
             res = c
     else :
           if xyz == 1:   res = Cvalue(c,1, n-1) + Cvalue(c,2, n-1)
           elif xyz == 2: res = Cvalue(c,2, n-1) + Cvalue(c,3, n-1)
           elif xyz == 3: res = Cvalue(c,3, n-1) + Cvalue(c,1, n-1)
           else:  print("Error\n")
     return res

def XYZvalues(x0, y0, z0, n):
    x = Cvalue(x0,1, n)
    y = Cvalue(y0,2, n)
    z = Cvalue(z0,3, n)
    return (x, y , z)

print(XYZvalues(1,1,1, 3))

迭代多个方程

3 个答案: