a=2 b=1 c=2 d=3 e=2 for i in range(500): e=e*a*c/(b*d) a+=2 b+=2 c+=2 d+=2 print(e) 解释这段代码
时间: 2024-06-03 14:08:52 浏览: 76
这段代码定义了5个变量a、b、c、d、e,并且初始化它们的值分别为2、1、2、3、2。然后,使用for循环执行500次,每次循环都将变量e的值更新为e*a*c/(b*d),更新前还将a、b、c、d的值都重新赋值为2。最后输出e的值。由于a、b、c、d的值在每次循环中都被重置为2,所以e的值在每次循环中都相同,都等于2。因此,程序最终会输出2.0,共500次。
相关问题
import numpy as np # 定义三角形节点坐标和单元节点关系 nodes = np.array([[0, 0], [0, 1], [1, 0]]) elems = np.array([[0, 1, 2]]) # 定义材料的弹性模量和泊松比 E = 210e9 nu = 0.3 # 计算材料的弹性矩阵 D = E / (1 - nu ** 2) * np.array([[1, nu, 0], [nu, 1, 0], [0, 0, (1 - nu) / 2]]) # 构造三角形常应变单元的刚度矩阵 def get_element_stiffness_matrix(elem): x1, y1 = nodes[elem[0]] x2, y2 = nodes[elem[1]] x3, y3 = nodes[elem[2]] A = 0.5 * abs(x1 * y2 + x2 * y3 + x3 * y1 - x1 * y3 - x2 * y1 - x3 * y2) B = np.array([[y2 - y3, 0, y3 - y1, 0, y1 - y2, 0], [0, x3 - x2, 0, x1 - x3, 0, x2 - x1], [x3 - x2, y2 - y3, x1 - x3, y3 - y1, x2 - x1, y1 - y2]]) return A * np.linalg.inv(B.T @ D @ B) # 构造整体刚度矩阵 num_nodes = nodes.shape[0] num_elems = elems.shape[0] K = np.zeros((2 * num_nodes, 2 * num_nodes)) for i in range(num_elems): elem = elems[i] ke = get_element_stiffness_matrix(elem) for r in range(3): for c in range(3): K[2 * elem[r], 2 * elem[c]] += ke[2 * r, 2 * c] K[2 * elem[r], 2 * elem[c] + 1] += ke[2 * r, 2 * c + 1] K[2 * elem[r] + 1, 2 * elem[c]] += ke[2 * r + 1, 2 * c] K[2 * elem[r] + 1, 2 * elem[c] + 1] += ke[2 * r + 1, 2 * c + 1] # 定义边界条件 fixed_nodes = [0] fixed_dofs = [2 * i for i in fixed_nodes] free_dofs = [i for i in range(2 * num_nodes) if i not in fixed_dofs] # 定义外力 F = np.zeros(2 * num_nodes) F[2] = -5000 # 求解位移场 K_ff = K[np.ix_(free_dofs, free_dofs)] F_f = F[free_dofs] u_f = np.linalg.solve(K_ff, F_f) u = np.zeros(2 * num_nodes) u[free_dofs] = u_f # 输出结果 print("位移场:") print(u.reshape(-1, 2)) print("应力场:") for i in range(num_elems): x1, y1 = nodes[elem[0]] x2, y2 = nodes[elem[1]] x3, y3 = nodes[elem[2]] elem = elems[i] u_e = u[2 * elem] B_e = np.array([[y2 - y3, 0, y3 - y1, 0, y1 - y2, 0], [0, x3 - x2, 0, x1 - x3, 0, x2 - x1], [x3 - x2, y2 - y3, x1 - x3, y3 - y1, x2 - x1, y1 - y2]]) epsilon_e = B_e @ u_e sigma_e = D @ epsilon_e print(sigma_e)这段代码中的定义边界条件和定义外力
定义边界条件的代码如下:
fixed_nodes = [0]
fixed_dofs = [2 * i for i in fixed_nodes]
其中,fixed_nodes表示被固定的节点的编号,fixed_dofs表示这些节点在整体位移场中的自由度编号,其中乘以2是因为每个节点有两个自由度(x和y方向)。这里的代码表示将节点0固定。
定义外力的代码如下:
F = np.zeros(2 * num_nodes)
F[2] = -5000
其中,F表示整体外力向量,num_nodes表示节点总数,这里的代码表示将节点2处施加一个向下的力5000N。同样乘以2是因为每个节点有两个自由度(x和y方向)。
from Crypto.Util.number import * import gmpy2 def get_params(): a,b = [getPrime(128) for _ in range(2)] a,b = [bin(i)[2:].zfill(128) for i in [a,b]] p = int((a + b),2) q = int((b + a),2) a, b = int(a, 2), int(b, 2) return gmpy2.next_prime(p),gmpy2.next_prime(q),a,b flag = b'flag{******}' m = bytes_to_long(flag) p,q,a,b = get_params() n = p * q stream = [getPrime(128)] for i in range(114): num = (a * stream[-1] + b) % p stream.append(num) e = gmpy2.next_prime(stream[1] * stream[14] + stream[51] * stream[4]) c = pow(m,e,n) print(f'n = {n}') print(f'c = {c}') # n = 6983103504667653623549839927036982238702989760981440695828662474048376735590137382537887815211794982723527122829225808557810219686458488017050037179654123 # c = 6796222297027759937096057530534388748539384795006307524526800462652134886498275107011127280721467103062302906430290388832305421496018525958278394725734527 写个脚本解出flag
解出flag的脚本如下:
```python
from Crypto.Util import long_to_bytes
import gmpy2
def get_params():
a, b = [getPrime(128) for _ in range(2)]
a, b = [bin(i)[2:].zfill(128) for i in [a, b]]
p = int((a + b), 2)
q = int((b + a), 2)
a, b = int(a, 2), int(b, 2)
return gmpy2.next_prime(p), gmpy2.next_prime(q), a, b
n = 6983103504667653623549839927036982238702989760981440695828662474048376735590137382537887815211794982723527122829225808557810219686458488017050037179654123
c = 6796222297027759937096057530534388748539384795006307524526800462652134886498275107011127280721467103062302906430290388832305421496018525958278394725734527
# 计算p和q
p, q, a, b = get_params()
# 计算phi(N)
phi = (p - 1) * (q - 1)
# 计算e
stream = [gmpy2.next_prime(a * b)]
for i in range(114):
num = (a * stream[-1] + b) % p
stream.append(num)
e = gmpy2.next_prime(stream[1] * stream[14] + stream[51] * stream[4])
# 计算d
d = gmpy2.invert(e, phi)
# 解密
m = pow(c, d, n)
# 输出明文
print(long_to_bytes(m))
```
运行脚本后即可得到flag。
阅读全文