#求圆的半径跟周长 圆周率3.14
r = int(input('Please enter the radius>>>'))
c = 1
while c:
arce = 3.14*r*r
perimeter = 2*3.14*r
c -= 1
print('The area is:', arce)
print('The perimeter is:', perimeter)
#输入两个数 比较大小 升序打印
c = 1
while c:
c -= 1
one = input('The first number:')
two = input('The second number')
if one > two:
print(one, two)
else:
print(two, one)
输入若干个整数 打印最大值 如果为0就退出
m = int(input('Please one number'))
while 1:
c = int(input('Please enter the number'))
if c == 0:
break
else:
n = c
if n > m :
m = n
print('Max=',m)
输入n个数 求每次的算数平均数
n = 0
m = 0
for i in range(5):
nums = int(input('Please a number'))
if nums == 0:
break
n += 1
m += nums
avg = m/n
print(avg)
打印边长为n的正方形
n = int(input('Please output side length'))
for i in range(n-1):
if i == 0:
print('*'*n)
else:
print('*'+' '*(n-2)+"*")
print（'*'*n)
求100以内所有奇数和
s = 0
for i in range(1,101,2):
s = s+i
print(s)
判断学生成绩
results = int(input('Please enter your score:'))
if results >= 70:
if 90 > results >= 80:
print('B')
elif results >= 90:
print('A')
else:
print('C')
else:
if results >= 60:
print('D')
else:
print('E')
求1到5的阶乘
k = int(input('Please enter a number:'))
nums = 1
count = 0
for i in range(1,k+1):
nums *= i
count += nums
print('阶乘=',nums)
print('阶乘和=',count)
给一个数 判断是否为质数
c = int(input('Please enter a number：'))
if c == 2:
print(c, 'Yes')
elif c % 2 == 0:
print(c, 'No')
if c % 2 != 0:
i = 2
while i < c:`在这里插入代码片`
if c % i == 0:
print(c,'No')
break
i += 1
else:
print(c,'Yes')
打印菱形
n = int(input('Please a number'))
if n % 2 == 0:
n += 1
for i in range(-int(n/2),int(n/2+1)):
print(' '*abs(i) + '*' * (n-2*abs(i)))
打印对顶三角形
n =7
e = n//2
for i in range(-e, n-e):
line = '*' * (2 * abs(i) + 1)
print('{:^{}}'.format(line,n))
打印闪电
n = 7
e = n // 2
for i in range(-e, e+ 1):
if i < 0:
print(' ' * (-i) + '*' * (4+i))
elif i == 0:
print('*' * n)
else:
print(' '*3 + '*'*(4-i))
解决猴子吃桃问题
peach = 1
for _ in range(9):
peach = (peach + 1) * 2
print(peach)
打印99乘法表
for i in range(1,10):
for j in range(1,i+1):
print('{}*{}={}{}'.format(j,i,i*j,' '*3 if j >1 and i*j <10 else ' '*2),end = ' \n' if j == i else '')
打印100以内的斐波那契数列
m = 0
n = 1
print(1)
while 1:
sum = m + n
if sum >100:
break
m = n
n = sum
print(sum)
打印斐波那契第101项
# m = 0 a = 0
# n = 1 b = 1
# sum = 0 count = 1
# for i in range(1,101): while 1:
# sum = m + n c = a + b
# m = n if count == 100:
# n = sum print(c)
# print(sum) break
# count += 1
# a, b = b, c
求10万以内的素数 基本版本
print(2)
for i in range(3,1000,2):
if i > 10 and i % 5 ==0:
continue
else:
for j in range(2,int(i**0.5+1)):
if i % j == 0:
break
else:
print(i)
求10万以内的素数
开方加6n+1 6n-1 （孪生定理）算法
start = datetime.datetime.now()
n = 100000
count = 3
x = 7
step = 4
while x < n:
if x % 5 != 0:
for i in range(3,int(x**5)+1,2):
if x % i == 0:
break
else:
count += 1
x += step
step = 4 if step == 2 else 2
delta = (datetime.datetime.now() - start).total_seconds()
print(count)
print(delta)
杨辉三角 补0法
l = int(input('Please one number:'))
m = [1]
n = []
s = []
for i in range(0,l): #循环用户输入的次数 这是列
n = m.copy() #copy一份m 给n
n.append(0) #在n后面追加一个0 目前n = [1，0]
m.clear() #清空一下m 每次都要清空
for j in range(i+1): #这里循环到用户输入+1的地方 这是行
m.append(n[j-1] + n[j]) #第一次进来用n的负1索引对应的数加上索引为0的数追加到m这个空列表里
print(m) #凑齐一行列表才打印
杨辉三角 单行实现
n = 9
line = [1] * n # 一次性开辟好于次次追加
print(line)
for i in range(n): # i
offset = n - i
swap = 1
for j in range(1, i//2+1): # i=3 range(1, 2) # 1
#val = swap + line[j] #line[j-1] + line[j]
#swap = line[j] # 3
#line[j] = val # 4
line[j], swap = swap + line[j], line[j] #等价上面三行代码
if i != 2*j:
line[-j-offset] = line[j]
print(line[:i+1])
