Python3语法入门题题解 & AcWing Django框架课

Python3语法入门题题解 & AcWing Django框架课

AcWing 608. 差

输入abcd，输出ab-cd

# 循环输入
a,b,c,d = (int(input()) for i in range(4))
print("DIFERENCA =",a*b-c*d)

  

 

  
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AcWing 665. 倍数

输入ab，判断是否有倍数关系

# split输入
a, b = map(int, input().split()) # split按空格和\n分割字符串, map转int
if b % a == 0 or a % b == 0:
    print("Sao Multiplos")
else:
    print("Nao sao Multiplos")

  

 

  
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AcWing 670. 动物

给出一个json数据，求输出某个叶节点的值

# 字典嵌套
t1={"carnivoro":"aguia","onivoro":"pomba"}
t2={"onivoro":"homem","herbivoro":"vaca"}
t3={"hematofago":"pulga","herbivoro":"lagarta"}
t4={"hematofago":"sanguessuga","onivoro":"minhoca"}
r1={"ave":t1,"mamifero":t2}
r2={"inseto":t3,"anelideo":t4}
s={"vertebrado":r1,"invertebrado":r2} 
a,b,c=input(),input(),input() # input每次读入一行
print(s[a][b][c])


  

 

  
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AcWing 660. 零食

给出 5 种零食的价目表。
求某种零食x数量为y时需要的钱。

# list输入
xy=list(map(int,input().split())) # map返回迭代器,list转为列表
print(f'Total: R$ {(4.00,4.50,5.00,2.00,1.50)[xy[0]-1]*xy[1]:.2f}')

  

 

  
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AcWing 760. 字符串长度

给出一行含空格的字符串，求长度

print(len(input()))

  

 

  
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AcWing 721. 递增序列

对于输入的整数n，输出1,2,3,n的序列
n=0时结束程序

n = int(input())
while n:
    for i in range (n):
        print(i+1, end = " ")
    print()
    n = int(input())

  

 

  
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AcWing 726. 质数

输入n个数，判断是不是质数

import math
n = int(input())
for i in range(n):
    x = int(input())
    s = math.floor(math.sqrt(x))
    p = True
    for j in range(2, s+1):
        if x%j==0 : 
            p = False
            break
    if p:
        print(x, "is prime")
    else:
        print(x, "is not prime")

  

 

  
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AcWing 745. 数组的右上半部分

给出一个12*12的矩阵，求右上半部分的平均值和元素和

op = input()
a = []
for i in range(12):
    a.append(list(map(float, input().split())))
res = 0
for i in range(12):
    for j in range(i+1, 12):
        res += a[i][j]
if(op=='M'): res/=66
print("%.1f"%(res))

  

 

  
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AcWing 756. 蛇形矩阵

给出n和m，将1-nm的数字按照回字蛇形填充至矩阵中

nm = input().split()
n = int(nm[0])
m = int(nm[1])
dx = [0,1,0,-1] # 按顺时针顺序走
dy = [1,0,-1,0]
res = [[0]*m for _ in range(n)] # 初始化数组
x,y,a,b,d = 0,0,0,0,0
for k in range(1,n*m+1):
    res[x][y] = k
    a=x+dx[d]  #每次走一步
    b=y+dy[d]
    if a<0 or a>=n or b<0 or b>=m or res[a][b]!=0: #判断合法
        d=(d+1)%4
        a=x+dx[d]
        b=y+dy[d]
    x = a
    y = b
for i in range(n):
    for j in range(m):
        print(f"{res[i][j]} ", end="")
    print()


  

 

  
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AcWing 823. 排列

给出n，求1-n的全排列

def dfs(cur):
    if cur > n:
        for i in range(1,n+1):
            print(a[i], end=" ")
        print()
    else :
        for i in range(1,n+1):
            if(vis[i] == False):
                vis[i] = True
                a[cur] = i
                dfs(cur+1)
                vis[i] = False

n = int(input()) #输入
a = [0]*(n+1)    #声明数组
vis = [False]*(n+1)
dfs(1)

  

 

  
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文章来源: gwj1314.blog.csdn.net，作者：小哈里，版权归原作者所有，如需转载，请联系作者。

原文链接：gwj1314.blog.csdn.net/article/details/125474520