import random
a=[0,0,0,0,0,0]
for i in range(6):
x = random.randint(1, 10)
if i % 2 == 0:
a[i] = 2 * x + 1
elif x % 2 == 0:
a[i] = x // 2
else:
a[i] = x - 1
执行该程序段后,a[0]~a[5]各元素可能的值是( )
A | 00001 | B | 00010 | C | 00011 | D | 00100 | E | 00101 | F | 00110 | G | 00111 |
H | 01000 | I | 01001 | J | 01010 | K | 01011 | L | 01100 | M | 01101 | N | 01110 |
O | 01111 | P | 10000 | Q | 10001 | R | 10010 | S | 10011 | T | 10100 | U | 10101 |
V | 10110 | W | 10111 | X | 11000 | Y | 11001 | Z | 11010 |
小明使用该编码对由大写字母组成的明文字符串进行加密,加密算法如下:
l依次将明文中每个字符转换为其对应的二进制编码;
l依次将密钥中每个字符转换为其对应的二进制编码;
l依次取出密钥的每个二进制位与原文的二进制位进行异或运算(若密钥长度不够,则循环重复使用),得到的结果即为密文编码;
l二进制位异或运算原则:11=0,00=0,10=1,01=1
例如,明文:HELLO,密钥:ZHE,则按上述方式进行加密:
明文字符 | H | E | L | L | O |
密钥字符 | Z | H | E | Z | H |
明文编码 | 01000 | 00101 | 01100 | 01100 | 01111 |
密钥编码 | 11010 | 01000 | 00101 | 11010 | 01000 |
密文编码 | 10010 | 01101 | 01001 | 10110 | 00111 |
def ctob(c): # 将一个字符转换为其对应的5位二进制编码
n =
ans = ""
for i in range(5):
r =
n = n // 2
ans = str(r) + ans
return ans
def xor(s1, s2): # 将二进制数s1和s2进行异或运算
ans = ""
for i in range(len(s1)):
if :
ans += "0"
else:
ans += "1"
return ans
s = input("请输入明文(大写字母):")
key = input("请输入密钥(大写字母):")
ans = ""
for i in range(len(s)):
s1 = ctob(s[i])
k =
s2 = ctob(key[k])
b = xor(s1, s2)
ans = ans + b
print("密文编码为:", ans)
①16/len("ab")**2
②abs(round(-1.7)*2)%8
③chr(ord("0")+4)
④int(str(1010+90)[1:4])//25
item={"竞技类":["铁人三项","电子竞技","霹雳舞"],"球类":["足球","篮球","乒乓球"],
"对抗性":["拳击","跆拳道","卡巴迪","击剑"],"水上":["跳水","龙舟","帆船"]}
print(item["对抗性"][2][::-1])
执行该程序段后,输出的结果是( )
s=1
for i in range(_):
s=s+i print(s)
import pandas as pd
s1=pd.Series(range(1,12,3))
print(s1)
该程序段运行后输出结果为( )
p = "Call-168"
c = " "
for ch in p:
if ch >= "0" and ch <= "9":
c += str(9-int(ch))
elif ch >= "a" and ch <= "z":
c += chr(ord(ch) - ord("a")+ord("A"))
else:
c += ch
print(c)
程序运行后,输出的结果是( )
s = "A-b2c " ; ch = ""
list=[4 , 3 , 1 , 5 ,2]
for i in range(1en( s)):
if "A " <= s [i] <= "Z " :
ch = ch + chr (ord(s [i]) + list[i])
elif "a " <= s [i] <= "z " :
ch = ch + chr (ord(s [i]) - list[i])
else:
ch= s [i]+ch
print(ch)
该程序运行后,输出的结果是( )