Common operators of 006:python

Notes

3 divisions in 1.python: traditional division, Precision division, floor removal.

Traditional Division : If it is an integer except that the law performs a floor addition, if it is a floating-point number, the law performs precise division.

>>>1/2  0  >>>1.0/2.0  0.5  

Exact Division : Division always returns the real quotient, regardless of whether the operand is shaped or floating. This can be done by executing from the future Import Division directive.

>>>fromimport division  >>>1/2  0.5  >>>1.0/2.0  0.5  

floor except ://Division Regardless of the numeric type of the operand, the fractional part is always removed, returning the nearest number in the sequence of numbers that is smaller than the true quotient.

>>>1//2  0  >>>1.0//2  0  >>>-1//2.0  -1  

2. Precedence of Operators: Power arithmetic > Sign > Arithmetic operators > Comparison operators > Logical operators.

Test questions

1. Please say the answer at the fastest speed: not 1 or 0 and 1 or 3 and 4 or 5 and 6 or 7 and 8 and 9.
Answer: 4

The precedence of not OR and is different: not > and > or

We enclose them in parentheses according to priority: (not 1) or (0 and 1) or (3 and 4) or (5 and 6) or (7 and 8 and 9)
= = 0 or 0 or 4 or 6 or 9
= = 4
Short Circuit logic:
If A and B: #如果a是false, then skip B's judgment and result directly false.
If A or B: #如果a为true, then skip B's judgment, direct true

2. Please write a program to print out all the odd numbers of 0~100.
For:

i = 0while i < 101:    if i %2 == 1:        print(i)        i += 1    else:        i += 1

Modify according to the answer:

i = 0while i <= 100:    if i % 2 != 0:        print(i, end=‘ ‘)        i += 1    else:        i += 1

3. Einstein had such an interesting math problem: there is a long ladder, if each step on the 2 order, the last remaining 1 orders, if each step 3 order, the last remaining 2 orders, if each step 5 order, the last 4 orders, if each step 6 order, finally left 5 order;

Title: Please program how many steps to solve the ladder?
For:

i = 0while 1:    if i%2==1 and i%3==2 and i%5==4 and i%6==5 and i%7==0:        print(i)        break    else:        i += 1

Modify according to the answer:

x = 7i = 1flag = 0while i <= 100:    if (x%2 == 1) and (x%3 == 2) and (x%5 == 4) and (x%6==5):        flag = 1    else:        x = 7 * (i+1) # 根据题意，x一定是7的整数倍，所以每次乘以7    i += 1if flag == 1:    print(‘阶梯数是：‘, x)else:    print(‘在程序限定的范围内找不到答案！‘)

Common operators of 006:python