python练习

发表于 更新于 分类于 本文字数： 8.6k 阅读时长 ≈ 8 分钟

打印图形

打印锥形

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
#给定一个数字和一个符号，打出锥形
import math
l = input().split()
n  = math.floor(math.sqrt((int(l[0])+1)/2))

for i in range(-n,n+1):
    if i < 0 :
        x = -i
        print(" "*(n-x)+l[1]*(2*x-1))
    elif i>1:
        x = i
        print(" "*(n-x)+l[1]*(2*x-1))
    else:
        print(end="")
z = n**2*2-1
print(int(l[0])-z)
31 *
*******
 *****
  ***
   *
  ***
 *****
*******
0

1
2
3
4
5
6
#坐标法打印锥形
n = 7
e = 7//2
for i in range(-e,e+1):
    x = -i if i < 0 else i
    print(" "*(e-x)+ "*" * (2*x+1))
*******
 *****
  ***
   *
  ***
 *****
*******

打印菱形

1
2
3
4
5
6
7
8
#打印菱形
n = 7
e = -(n//2)
for i in range(e,n+e):
    if i<=0:
        print(" "*abs(i)+"*"*(n+2*i))
    else :
        print(" "*i+"*"*(n-2*i))
   *
  ***
 *****
*******
 *****
  ***
   *

打印闪电

1
2
3
4
5
6
7
8
9
10
#打印闪电
n = 7
e = n//2
for i in range(-e,e+1):
    if i<0:
        print(" "*abs(i)+ "*" * (e+i+1))
    elif i == 0:
        print("*"*n)
    else:
        print(" "*3+"*"*(e-i+1))
   *
  **
 ***
*******
   ***
   **
   *

打印正方形

1
2
3
4
5
6
7
8
#打印一个边长为n的正方形
var =  int(input(">>"))
for i in range(var):
    if i ==0 or i == var-1:
        print("*"*var)
    else:
        print("*" + " "*(var-2) + "*")
    
>>4
****
*  *
*  *
****

1
2
3
4
5
6
n = 4
for i  in range(n):
    if i % (n-1)==0:
        print("*"*n)
    else:
        print("*"+" "*(n-2)+"*")
****
*  *
*  *
****

1
2
3
4
5
6
7
8
#打印一个边长为n的正方形
n =  int(input(">>"))
e = -(n//2)
for i in range(e,n+e):
    if i == e or i == n+e-1:
        print ("*"*n)
    else :
        print("*"+" "*(n-2)+"*")
>>6
******
*    *
*    *
*    *
*    *
******

1
2
3
4
5
6
7
8
9
10
11
12
13
#统计你输入的每个数字有多少个，按照升序输出
l = input().split()
l = int(l[0])
num = list()
num1 = [0]*10
while l>0:
    x = l % 10
    num1[x] += 1
    num.append(x)
    l //=10
for i in range(10):
    if num1[i] !=0:
        print("{}:{}".format(i,num1[i]))
1231013
0:1
1:3
2:1
3:2

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
#读数字
zz = input().split()
a = int(zz[0])
l = ['ling','yi','er','san','si','wu','liu','qi','ba','jiu','fu']
i = 0
li =list()
x = a
y = abs(a)
while abs(x)>0:
    x = abs(x)//10
    i += 1
for j in range(i):
    m = 10**(i-1)
    y = y// m
    li.append(y)
    i -= 1
    y = abs(a) % m

length = len(li)
if a>0:
    for i in li:
        if int(i)<length:
            print("{}".format(str(l[i])),end=" ")
        else:
            print("{}".format(str(l[i])),end="")
else:
    print(str(l[10]),end=" ")
    for i in li:
        if int(i)<length:
            print("{}".format(str(l[i])),end=" ")
        else:
            print("{}".format(str(l[i])),end="")
-123
fu yi er san

三个数排序

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
#只用判断，三个数排序
l = list()
for i in range(3):
    l.append(int(input(">>")))
if l[0] > l[1]:
    if l[0] > l[2]:
        if l[1]>l[2]:
            print(l[0],l[1],l[2])
        else:
            print(l[0],l[2],l[1])
    else:
        print(l[2],l[1],l[0])
else :
    if l[0] > l[2]:
        print(l[1],l[0],l[2])
    else:
        if  l[1]<l[2]:
            print(l[2],l[1],l[0])
        else:
            print(l[1],l[2],l[0])
    
>>1
>>3
>>2
3 2 1

1
2
3
4
5
6
7
8
9
10
11
#使用max函数，三个数排序
l = list()
for i in range(3):
    l.append(int(input(">>")))
while l:
    if len(l) == 1:
        print(l[0])
        break
    m = max(l)
    print(m)
    l.remove(max(l))    
>>2
>>3
>>1
3
2
1

1
2
3
4
5
6
#使用sort（）函数，三个数排序
l = list()
for i in range(3):
    l.append(int(input(">>")))
l.sort()
l
>>1
>>4
>>2

[1, 2, 4]

冒泡法排序（重要）

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
#优化后的冒泡法（重要）
#l=[1,9,8,5,6,7,4,3,2]
l=[1,2,3,4,5,6,7,9,8]
length = len(num)
count = 0
count_swap = 0

for i in range(length):
    flag = False
    for j in range(length-i-1):
        count += 1
        if l[j] > l[j+1]:
            tmp = l[j]
            l[j] = l[j+1]
            l[j+1] = tmp
            flag = True
            count_swap += 1
    if not flag:
        break
print(l,count,count_swap)
            
[1, 2, 3, 4, 5, 6, 7, 8, 9] 15 1
  • 优化点：可以设定一个标记判断此轮是否有数据交换发生，如果没有发生交换，可以结束排序，如果发生交换，则进行下一轮
  • 最差的排序情况是，初始顺序和目标顺序完全相反，遍历次数1，…,n-1之和n*（n-）/2
  • 最好的排序情况是，初始顺序与目标顺序完全相同，遍历次数n-1
  • 时间复杂度O（n^2）

二分法查找

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
#二分法查找
def banery_serch(list,item):
    low = 0
    high = len(list)-1
    while low <= high:
        mid = (low+high)//2
        giss = list[mid]
        if item == mid:
            print(Ture)
        elif item > mid:
            high = mid - 1
        else:
            low = mid +1
my_list = [1,2,3,5,2,6,4,7,9]

给定一个数，输入为几位数，并依次输出

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
 #给定一个数，输入为几位数
num = int(input(">>"))
i = 0
flag = False
while num > 0:
    a = num%10
    print("a",a,end=" ")
    num = num//10
    
#给定一个数字判断有几位
num = int(input(">>"))
i = 1
while num >= 10:
    num = num/10
    i += 1
print("i",i)
>>1234
a 4 a 3 a 2 a 1 

1
2
3
4
5
6
7
8
9
10
11
12
13
 #给定一个数，并依此输出\n
x = '02300'
x = int(x)
n = 5
w = 10000
flag = False
for i in range(n):
    y = x // w
    if flag or y:
        flag = True    
        print(y,end=" ")
    x = x % w
    w = w //10
2 3 0 0 

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
c = int('00211')
w = 10000
length = 5
flag = False
while w:
    t = c //w
    if flag:
        print(t,end=" ")
    else:
        if t:
            print(t,end=" ")
            flag =True
        else:
            length -=1
    c %= w
    w //=10
2 1 1

求均值

1
2
3
4
5
6
7
8
9
10
11
a = int(input(">>")) 
l = list() 
num = 0 
for i in range(a): 
    x = int(input('>>')) 
    if x == None: 
        break 
    else: 
        l.append(x) 
        num = num+l[i] 
        print(num/len(l))
>>3
>>1
1.0
>>3
2.0
>>4
2.6666666666666665

打印乘法表

打印倒乘法表

1
2
3
4
5
6
#打印倒乘法表
for i in range(1,10):
    print("\t"*(i-1),end="")
    for j in range(i,10):
        print ("{}x{}={:<{}}".format(i,j,i*j,2 if j<2 else 3),end=" ")
    print()
1x1=1  1x2=2   1x3=3   1x4=4   1x5=5   1x6=6   1x7=7   1x8=8   1x9=9   
       2x2=4   2x3=6   2x4=8   2x5=10  2x6=12  2x7=14  2x8=16  2x9=18  
               3x3=9   3x4=12  3x5=15  3x6=18  3x7=21  3x8=24  3x9=27  
                       4x4=16  4x5=20  4x6=24  4x7=28  4x8=32  4x9=36  
                               5x5=25  5x6=30  5x7=35  5x8=40  5x9=45  
                                       6x6=36  6x7=42  6x8=48  6x9=54  
                                               7x7=49  7x8=56  7x9=63  
                                                       8x8=64  8x9=72  
                                                               9x9=81

打印正乘法表

1
2
3
4
5
#打印九九乘法表：
for i in range(1,10):
    for j in range(1,i+1):
        print("{}x{}={}\t".format(j,i,i*j),end=" ")
    print()
1x1=1     
1x2=2     2x2=4     
1x3=3     2x3=6     3x3=9     
1x4=4     2x4=8     3x4=12     4x4=16     
1x5=5     2x5=10     3x5=15     4x5=20     5x5=25     
1x6=6     2x6=12     3x6=18     4x6=24     5x6=30     6x6=36     
1x7=7     2x7=14     3x7=21     4x7=28     5x7=35     6x7=42     7x7=49     
1x8=8     2x8=16     3x8=24     4x8=32     5x8=40     6x8=48     7x8=56     8x8=64     
1x9=9     2x9=18     3x9=27     4x9=36     5x9=45     6x9=54     7x9=63     8x9=72     9x9=81     

1
2
3
4
5
#打印九九乘法表：
for i in range(1,10):
    for j in range(1,i+1):
        print(j,"x",i,"=",j*i," ",end=" ")
    print()
1 x 1 = 1   
1 x 2 = 2   2 x 2 = 4   
1 x 3 = 3   2 x 3 = 6   3 x 3 = 9   
1 x 4 = 4   2 x 4 = 8   3 x 4 = 12   4 x 4 = 16   
1 x 5 = 5   2 x 5 = 10   3 x 5 = 15   4 x 5 = 20   5 x 5 = 25   
1 x 6 = 6   2 x 6 = 12   3 x 6 = 18   4 x 6 = 24   5 x 6 = 30   6 x 6 = 36   
1 x 7 = 7   2 x 7 = 14   3 x 7 = 21   4 x 7 = 28   5 x 7 = 35   6 x 7 = 42   7 x 7 = 49   
1 x 8 = 8   2 x 8 = 16   3 x 8 = 24   4 x 8 = 32   5 x 8 = 40   6 x 8 = 48   7 x 8 = 56   8 x 8 = 64   
1 x 9 = 9   2 x 9 = 18   3 x 9 = 27   4 x 9 = 36   5 x 9 = 45   6 x 9 = 54   7 x 9 = 63   8 x 9 = 72   9 x 9 = 81

​

一到五阶乘之和

1
2
3
4
5
6
7
8
#一到五阶乘之和
a= 1
sum = 0
for i in range(1,6):
    a *= i
    sum += a
    i += 1
print("sum =",sum )
1
2
3
4
5
#100内的所有奇数之和
a = 0
for i in range (1,100,2):
    a = a+i
print("a=",a)
a= 2500

1
2
3
4
5
6
#100内的所有奇数之和
a = 0
for i in range(1,100,2):
    if not i%2 == 0:
        a += i
print("a=",a)

斐波那契数列

打印100以内的斐波那契数列

1
2
3
4
5
6
# 打印100以内的斐波那契数列
i = 0
j = 1
while i < 100:
    print(i,end=" ")
    i,j=j,i+j
0 1 1 2 3 5 8 13 21 34 55 89 

1
2
3
4
5
6
7
8
# 打印100以内的斐波那契数列
i = 0
j = 1
while i<=100:
    print (i,end=" ")
    k = i + j
    i = j
    j = k
0 1 1 2 3 5 8 13 21 34 55 89

求斐波那契数列第101项

1
2
3
4
5
6
7
8
9
#求斐波那契数列第101项
i = 0
j = 1
m = 1
while m<=101:
    i, j = j, i + j
    m += 1
    
print(i) 
573147844013817084101

求素数

寻常法

1
2
3
4
5
6
7
8
9
10
11
12
13
#求10万内的所有素数
import math
n = 100000
count = 1
for i in range(3,n,2):
    if i > 10 and i % 5 ==0:
        continue  #所有大于10 的素数中，个位数只有1，3，7，9，所以此处筛出5结尾的
    for j in range(3,int(math.sqrt(i)+1),2):
        if i % j ==0:
            break
    else:
        count += 1
print (count)
9592

利用素数列表，空间换时间

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
#存储质数
#合数一定可以分解为几个质数的乘积，2是质数
#质数一定不能整除1和本身之内的整数
#使用质数来取模比利用基数取模计算次数更少
import math
n = 100000
count = 1
primenumber = [2]
for i in range(3,n,2):
    flag = False
    edge = math.sqrt(i)
    for j in primenumber:
        if i % j == 0:
            flag = True
            break
        if j > edge:
            break
    if not flag:
        count += 1
        primenumber.append(i)
print(count)

利用孪生素数

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
#大于3 的素数只有6N-1和6N+1两种形式，如果6N-1和6N+1都是素数便叫作孪生素数
import math
n = 100
count = 2
primenumber = [2,3]
step = 2
x = 5
while x < n:
	for i in range(3,int(math.sqrt(x)+1),2):
		if x % i == 0:
			break
	else:
		count += 1
		primenumber.append(x)
	x += step
	step = 4 if step == 2 else 2
print(count)
print(primenumber)

埃式筛素数

1
2
3
4
5
6
7
8
9
10
11
12
#埃式筛素数
def eratosthenes(n):
    IsPrime = [True] * (n + 1)
    IsPrime[1] = False
    for i in range(2, int(n ** 0.5) + 1):
        if IsPrime[i]:
            for j in range(i * 2, n + 1, i):
                IsPrime[j] = False
    return {x for x in range(2, n + 1) if IsPrime[x]}

if __name__ == "__main__":
    print (eratosthenes(100))
{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

杨辉三角

计算杨辉三角，普通法

1
2
3
4
5
6
7
8
9
10
#计算杨辉三角 普通法
triangle = [[1],[1,1]]
for i in range(2,6):
    swap = triangle[i-1]
    cul = [1]
    for j in range(i-1):
        cul.append(swap[j]+swap[j+1])
    cul.append(1)
    triangle.append(cul)
triangle

计算杨辉三角 补0法

1
2
3
4
5
6
7
8
9
10
#计算杨辉三角 补0法
triangle = [[1]]
n = 7
for i in range(1,n):
    swap = triangle[i-1]+[0]
    cul = [1]
    for j in range(len(swap)-1):
        cul.append(swap[j]+swap[j+1])
    triangle.append(cul)
print(triangle)

杨辉三角，对称法

1
2
3
4
5
6
7
8
9
10
11
12
#杨辉三角，对称法
n=6
triangle = [[1],[1,1]]
for i in range(2,n):
    tmp = triangle[-1]
    cul = [1] * (i+1)
    for j in range(i//2):
        cul[j+1] = tmp[j]+tmp[j+1]
        if i != 2j:	
            cul[-j-2] = cul[j+1]
    triangle.append(cul)
triangle

中点的确定
[1]
[1,1]
[1,2,1]
[1,3,3,1]
[1,4,6,4,1]
[1,5,10,10,5,1]
把整个杨辉三角看成一个左对齐的二维矩阵。
i == 2时，在第3行，中点的列索引j==1
i == 3时，在第4行，无中点
i == 4时，在第5行，中点的列索引j==2
得到以下规律，如果i==2j，则有中点

杨辉三角，单列表方法

1
2
3
4
5
6
7
8
9
10
11
12
13
#杨辉三角，单列表解决
n = 6
row = [1] * n
for i in range(n):
    z = 1
    offset = n - i
    for j in range(1,i//2+1):
        val = z + row[j]
        z = row[j]
        row[j] = val
        if i != 2*j:
            row[-j - offset] = val
    print(row[:i+1])
相关文章