A job
A. Short answer questions (a total of five questions, 100.0 points)
1
A plurality of x = 6+ 8j known Please write its mold, the real part and the imaginary part of the complex conjugate of the command, and write the results.
correct answer:
my answer:
x = 6 + 8j
print(abs(x))
print(x.real)
print(x.imag)
print(x.conjugate())
result:
10.0
6.0
8.0
(6-8j)
2
Known that a = [97, 98, 99,100], b = (97, 98, 99,100), c = { 'x': 97, 'y': 98, 'z': 99, 'q': 100}, d = {97, 98, 99,100}, is a [1], b [1], c [1], d [1] and c [ 'q']) operating results What are they? And explain the reasons.
correct answer:
my answer:
a= [97, 98, 99,100]
print(a[1])
b = (97, 98, 99,100)
print(b[1])
operation result:
98
98
The reason: a, is a list, b is a tuple. Lists, tuples supporting elements subscripting specified position, are numbered from 0, a [1], b [1] is the second output element, the output is 98, 98.
c = {‘x’:97, ‘y’:98, ‘z’:99,‘q’:100}
print(c[1])
operation result:
print(c[1])
KeyError: 1
The reason: c is a dictionary, dictionaries support subscript is key. Elemental form dictionary is "key: value." Error thrown when a keyword, which means using the keyword (key) mapping does not exist: the key is not in c 1, the output KeyError.
d = {97, 98, 99,100}
print(d[1])
operation result:
print(d[1])
TypeError: ‘set’ object is not subscriptable
Because:
d is a collection of the collection does not support the index, so TypeError, the type of error. A collection of commonly used in, not in determining whether the elements in the collection.
c = {‘x’:97, ‘y’:98, ‘z’:99,‘q’:100}
print (c [q])
Operating results are:
100
c [ 'q'] c, can be found in a dictionary key 'q', which corresponds to the value 100, the output is 100.
3
Run the following command to write the results and explain the reasons.
2 or 3
2 and 3
2 and 3 and True
0 and 3 and 1
0 and 3 or 1
0 and 3 or 1 or 4
0 or False and 1
correct answer:
my answer:
Operating results are:
2
3
True
0
1
1
False
Because:
Priority: ()> not> and> or, operators and, or do not necessarily return true or false, but rather to give a final calculated value of the expression.
在python中逻辑运算符or，x or y ,如果 x or y ，x为true则返回x,如果x 为false返回y值。因为一个为真则为真，所以返回x的值，如果x的值为假，那么or运算的结果取决于y,所以返回y值。逻辑运算符and,x and y ,若x为true则返回y值，若x为false则返回x值。因为x为true，and的运算不会结束，会继续看y的值，所以此时真假取决于y的值，所以x为真则返回y的值。如果x为假，那么and运算就结束运算过程，因为一个为假则为假，所以返回x的值。
2 or 3: 2,3 is true, or: a true calculation result is True, it returns to the first operation result true value of 2
2 and 3: 2,3 is true, and: two true value is True, the second number of the calculation result being true 3
2 and 3 and True: 2,3, True are true, 2 and 3 the operation result is 3; continue operation, 3 and true, the operation result is true
0 and 3 and 1: 0 is false, true 3, 3 0 and the operation result is 0; 0 and 1,0 is false, the end of the operation, the result is 0 (0 for any result of a logical AND operation is 0)
0 and 3 or 1: 0 and 3 the operation result is 0; 0 is false, 1 is true, the return value is 0 or 1 1 (1 ORed result with any number of 1)
0 and 3 or 1 or 4: 0 is false, 0 and 3 0 returns true value, 0 or 1, 1 is the return value is true, 1 or 4 end of the operation returns a value of 1
0 or False and 1: and superior operational level or, the first operation returns a value of False and 1 False (and: a false, false), then computing 0 or False, 0 is false, the operation result is False (or : two false and false, operation results for the second false number).
4
writing small programs, enter three sides of a triangle, the output of its area, and write the results.
correct answer:
my answer:
x = int (input ( "Enter a triangle side x:"))
y = int (input ( "Enter a side of the triangle y:"))
z = int (input ( "Enter a side of the triangle z:"))
c=(x+y+z)/2
s=(c*(c-x)(c-y)(c-z)) ** 0.5
print ( 'triangle area% 0.2f'% s)
result:
Please enter one side of the triangle x: 5
Please enter one side of the triangle y: 6
Please enter one side of the triangle z: 8
Triangle area of 14.98
5
writing small programs, two sets setA input and set B, respectively, the output of their intersection, union, symmetric difference and difference, and write the results.
correct answer:
my answer:
Arrows = {1,2,3,4}
setB={3,4,6,8}
a=((setA) & (setB))
b = ((silk) | (SETB))
c = ((silk) ^ (SETB))
d = ((SETA) - (SETB))
e = ((setB) - (arrow))
print ( "Intersection:% s"% a)
print ( "union:% s"% b)
print ( "symmetric difference:% s"% c)
print ( "A and B set difference:% s"% d)
print ( "B and A difference set:% s"% e)
The results are:
Intersection: {3, 4}
And set: {1, 2, 3, 4, 6, 8}
Symmetric difference: {1, 2, 6, 8}
The difference between A and B set: {1, 2}
B and A difference set: {8, 6}
