有两种方法可以破解维吉尼亚密码。一种方法使用强力字典攻击来尝试将字典文件中的每个单词作为维吉尼亚密钥,只有当该密钥是英语单词时才有效,如 RAVEN 或 DESK。第二种更复杂的方法是 19 世纪数学家查尔斯·巴贝奇使用的,即使密钥是一组随机的字母,如 VUWFE 或 PNFJ,它也能工作。在本章中,我们将使用这两种方法编写...
borrow = a->ob_digit[i] - b->ob_digit[i] - borrow; z->ob_digit[i] = borrow & PyLong_MASK; borrow >>= PyLong_SHIFT; borrow &= 1; /* Keep only one sign bit */ } for (; i < size_a; ++i) { borrow = a->ob_digit[i] - borrow; z->ob_digit[i] = borrow & PyL...
borrow = a->ob_digit[i] - b->ob_digit[i] - borrow; /* 第一轮:z->ob_digit[0] = 16 & 127 = 16 第二轮:z->ob_digit[1] = 3 & 127 = 3 */ z->ob_digit[i] = borrow & PyLong_MASK; /* 第一轮:borrow = 16 >> 7 = 0 第二轮:borrow = 3 >> 7 = 0 */ borrow >...
Inside the for loop, the program will pick two single-digit numbers to multiply. We’ll use these numbers to create a #Q: N × N = prompt for the user, where Q is the question number (1 to 10) and N are the two numbers to multiply. # Pick two random numbers: num1 = random....
Your program should be able to generate simple addition problems that involve adding two 2-digit integers (i.e., the numbers 10 through 99). The user should be asked for an answer to each problem. Your program should determine if the answer was correct or not, and give the user an appr...
10、“The rules for parsing an item key are very simple. If it starts with a digit, then it is treated as a number, otherwise it is used as a string.” 11、Within a replacement field, a colon (:) marks the start of the format specifier. ...
If the digit is odd, then you round up. That’s why 2.5 rounds down to 2 and 3.5 rounds up to 4.Note: Rounding ties to even is the rounding strategy recommended for floating-point numbers by the IEEE(Institute of Electrical and Electronics Engineers) because it helps limit the impact ...
Here are some things to keep in mind: If the variable raises an exception when called, the exception will be propagated, unless the exception has an attribute silent_variable_failure whose value is True. If the exception does have a silent_variable_failure attribute whose value is True, the ...
import unicodedata import re re_digit = re.compile(r'\d') sample = '1\xbc\xb2\u0969\u136b\u216b\u2466\u2480\u3285' for char in sample: print('U+%04x' % ord(char), char.center(6), 're_dig' if re_digit.match(char) else '-', 'isdig' if char.isdigit() else '-', 'is...
{ borrow = a->ob_digit[i] - borrow; z->ob_digit[i] = borrow & PyLong_MASK; borrow >>= PyLong_SHIFT; borrow &= 1; /* Keep only one sign bit */ } assert(borrow == 0); if (sign < 0) { Py_SET_SIZE(z, -Py_SIZE(z)); } return maybe_small_long(long_normalize(z))...