用ord和chr函数会很方便,特判z
- n=input()
- a=[]
- for x in n:
- if(x=='z'):
- a.append(100)
- else:
- a.append(ord(x)+4)
- for i in a:
- print(chr(i),end='')
- n=float(input())
- print("c="+"%.2f"%(5*(n-32)/9))
反向循环或reserve
- s=input()
- print(len(s))
-
- for i in s:
- print(i,end=" ")
- print("")
- for i in range(len(s)-1,-1,-1):
- print(s[i],end="")
大模拟,最后写的,能少罚时
- a=int(input())
- m=0
- #print(100000*0.1)
- #print(100000*0.075)
- #print(200000*0.05)
-
- if a<100000:
- m=a*0.1
- if a>100000 and a<=200000:
- m=10000+(a-100000)*0.075
- if a>200000 and a<=400000:
- m=17500+(a-200000)*0.05
- if a>400000 and a<=600000:
- m=27500+(a-400000)*0.03
- if a>600000 and a<=1000000:
- m=33500+(a-600000)*0.015
- if a>1000000:
- m=39500+(a-1000000)*0.01
-
- m=int(m)
- print(m)
- a,b,c=map(int,input().split())
- num=0
- for i in range(1,a+1):
- num+=i
- for i in range(1,b+1):
- num+=i**2
- for i in range(1,c+1):
- num+=1/i
- print("%.2f"%num)
这四个数背都背下来了 QAQ
- print("153")
- print("370")
- print("371")
- print("407")
- n=int(input())
- for i in range(2,n):
- lis=[]
- sum=0
- for j in range(1,i//2+1):
- if i%j==0:
- lis.append(j)
- sum+=j
- if sum==i:
- print(i,end="")
- print(" its factors are ",end="")
- for t in lis:
- print(t,end=" ")
- print("")
- n=int(input())
- sum=0
- a=2
- b=1
- for i in range(0,n):
- sum+=a/b
- tmp=a
- a=a+b
- b=tmp
- print("%.2f"%sum)
每弹跳一次减半
问总路径和最后高度
- m,n=map(int,input().split())
- sum=0
- for i in range(0,n):
- sum+=m+m/2
- m/=2
- sum=sum-m
- print("%.2f %.2f"%(m,sum))
筛质数
- n=int(input())
- for i in range(2,n):
- fl=0
- for j in range(2,i//2+1):
- if i%j==0:
- fl=1
- break
- if fl==0:
- print(i)