会员歌曲怎么办呢……

解决

恢复了 PRAM

哇，好多回复，我挨个搜搜去。
谢谢各位 !

@shinko
@lhbc
@widewing
@xspoco
@CEBBCAT

按照回复时间排序。

google+

Facetime ，女票的手机是大陆行货，没有 Facetime Audio ，不然微信语音通话也可以不用了。

一切遵循 Google Coding Style

@hzw MacOS Sierra 使用你给出的系统自带命令，显示无可用更新。
用了 mas ， 正常更新……

Valgrind 用不了了……
不知道有什么可以代替……
检查 c

自己顶一下

新人提问： PORT 是多少……

以下的 （ may be ） 是最终版本：

# Paviller CryptoSystem on SageMath
# Feb. 03, 2016

def getRandom ():
tmp = 0;
while (tmp == 0):
r = ZZ.random_element(2^(512 - 1), 2^512)
# random number 512 bits from 2^(512 - 1) to 2^215 - 1
if is_prime(r):
tmp = 1
return r

def getKeyList ():
LKey = []

# initialize prime number p and q
p = getRandom()
q = getRandom()
while (p == q):
p = getRandom()
LKey.append(p) #Lkey[0]
LKey.append(q) #Lkey[1]

lambdan = lcm(p - 1, q - 1)
LKey.append(lambdan) #Lkey[2]

n = p * q
LKey.append(n) #Lkey[3]

if (gcd(n, lambdan) != 1):
return false

g = n + 1
LKey.append(g) #Lkey[4]

# how it works for returning listKey1 & listKey2 at the same time?
return LKey


def getPubKey (LKey):
KPub = []
KPub.extend(LKey[3:5])
print("Public Keys: n and g")
return KPub

def getPriKey (LKey):
KPri = []
KPri.extend(LKey[0:3])
print("Private Keys: p, q and lamdba(n)")
return KPri

def Encoding (m, KPub):
n = KPub[0]
Zn = Integers(n)
Znn = Integers(n^2)
g = Znn(KPub[1])
r = 0
while (r == 0): # r non-null
r = Znn((Zn.random_element()))
c = Znn(g^m * r^n)
return c

def L(u, KPub):
n = KPub[0]
Znn = Integers(n^2)
return ((Znn(u).lift() - 1) / n)

def Decoding (c, KPub, KPri):
n = KPub[0]
Zn = Integers(n)
Znn = Integers(n^2)
g = Znn(KPub[1])
lambdan = Znn(KPri[2])

Pup = L((c^lambdan), KPub)
Pdown = L((g^lambdan), KPub)
m = Zn(Pup / Pdown)
return m

# fast test with:
# m = getRandom(); m # a number as clear message
# LKey = getKeyList(); KPub = getPubKey (LKey); KPri = getPriKey (LKey); c = Encoding (m, KPub); Decoding (c, KPub, KPri)

放在 GitHub 了： https://github.com/MrBowen/Paviller-on-SageMath.git

已解决：
完成版代码见维护末尾。
此代码为了节省运算，将 g 定义为 g = 1 + n 。

（新问题来了： mac 怎么打 “ ` ”，如果用 1 键左边的，我打出来是“ · ”）
```python

def getRandom ():
tmp = 0;
while (tmp == 0):
r = ZZ.random_element(2^(512 - 1), 2^512)
# random number 512 bits from 2^(512 - 1) to 2^215 - 1
if is_prime(r):
tmp = 1
return r

def getKeyList ():
LKey = []

# initialize prime number p and q
p = getRandom()
q = getRandom()
while (p == q):
p = getRandom()
LKey.append(p) #Lkey[0]
LKey.append(q) #Lkey[1]

lambdan = lcm(p - 1, q - 1)
LKey.append(lambdan) #Lkey[2]

n = p * q
LKey.append(n) #Lkey[3]

if (gcd(n, lambdan) != 1):
return false

g = n + 1
LKey.append(g) #Lkey[4]

# how it works with return listKey1, listKey2 ?
return LKey


def getPubKey (LKey):
KPub = []
KPub.extend(LKey[3:5])
return KPub

def getPriKey (LKey):
KPri = []
KPri.extend(LKey[0:3])
return KPri

def Encoding (m, KPub):
n = KPub[0]
Zn = Integers(n)
Znn = Integers(n^2)
g = Znn(KPub[1])
r = Znn(abs(ZZ.random_element()))
c = Znn(g^m * r^n)
return c

def L(u, KPub):
n = KPub[0]
Zn = Integers(n)
Znn = Integers(n^2)
return Zn((u - 1).lift() / n)

def Decoding (c, KPub, KPri):
n = KPub[0]
Zn = Integers(n)
Znn = Integers(n^2)
g = Znn(KPub[1])
lambdan = Znn(KPri[2])

Pup = L(pow(c, lambdan, n^2), KPub)
Pdown = L(pow(g, lambdan, n^2), KPub)
m = Zn(Pup / Pdown)
return m

自己顶一下……

为咩不能编辑了……
Zn 数学的表达是 Z/nZ
Zn^2 数学的表达是 Z/n^2Z

@764664 谢谢，现在去搜