Table of contents
1. Find the sum of numbers between 1 and 100 that are not divisible by 3
2. Given a positive integer N, find the sum of all prime numbers between 1 and N (inclusive)
3. Calculate PI (the formula is as follows: PI=4 (1-1/3+1/5-1/7+1/9-.....)
6. Merge two ordered arrays, and the merged array will still be an ordered list.
1. Find the sum of numbers between 1 and 100 that are not divisible by 3
sum = 0
for a in range (1, 101):
if a % 3 == 0:
continue
else:
sum += a
print(sum)
2. Given a positive integer N, find the sum of all prime numbers between 1 and N (inclusive)
def xy(x):
if x == 1:
return False
for i in range(2, x // 2 + 1):
if x % i == 0:
return False
return True
N = int(input('请输入一个正整数N：'))
sum = 0
for x in range(1, N + 1):
if xy(x):
sum += x
print(f'1到N(含)之间所有质数的总和为：{
3. Calculate PI (the formula is as follows: PI=4 (1-1/3+1/5-1/7+1/9-.....)
def PI():
n = 0
sum_PI = 0
for i in range(1, 10000, 2):
sum_PI += ((-1) ** n) * (1 / i)
n += 1
PI = 4 * sum_PI
return PI
print(f'PI = {PI()}')
4. Find a+aa+aaa+......+aaaaaaaaa=? Where a is a number from 1 to 9, and the number of terms must also be specified.
def sum(a, n):
sum_a = 0
for i in range(1, n + 1):
num = int(f'{a}' * i)
sum_a += num
return sum_a
a = int(input('请输入一个在区间[1,9]的正整数：'))
n = int(input('请输入指定的项数：'))
print(f'多项式的和为：{sum(a, n)}')
5. Find a number (function) within 10,000 that is divisible by 5 or 6, but not both at the same time.
def func():
for i in range(1, 10001):
if (i % 5 == 0 or i % 6 == 0 ):
if i % 5 == 0 and i % 6 == 0:
continue
print(i)
func()
6. Merge two ordered arrays, and the merged array will still be an ordered list.
arr1 = [1, 3, 4, 6, 10]
arr2 = [2, 5, 8, 11]
ans = arr1 + arr2
ans.sort()
print(ans)
7. Write a method to calculate the sum of all even-numbered elements in the list (note the return value)
arr = [1, 2, 3, 4, 6, 10, 11, 13]
sum = 0
for i in arr:
if i % 2 != 0:
continue
else:
sum += i
print(sum)
8. Given an array A of non-negative integers, place all even numbers in the array before the odd elements
def even_before_odd(ls = []):
for i in range(len(ls)):
if ls[i] % 2 != 0:
for j in range(i + 1, len(ls)):
if ls[j] % 2 == 0:
ls[i], ls[j] = ls[j], ls[i]
break
return ls
ls = [1, 2, 3, 4, 5, 6, 7, 8, 9]
print(f'原列表为：{ls}')
print(f'将偶数置于奇数前，列表变为：{even_before_odd(ls)}')