I spent one or two days writing this code. This small program is very interesting and useful, especially when you book the airline ticket and do not know how much time it'll cost during each phase.
For ...
read morea=89
b=113
print '%8d'%a
print 'x%7d'%b
c=b
print '-'*8
i=8
while c>0:
print '%*d'%(i,(c%10)*a) # * means how many digit, i determines it
c=c/10
i-=1
if i==7: # if b is 1 digit decimal, no more ------
print ...# A square joining together
class Solution(object):
def removeoneside(self,nums,side):
n=len(nums)
res = [[0 for j in range(side + 1)] for i in range(n + 1)]
# res[i][j] = max(res[i - 1][j],res[i-1][j-nums[i-1]]+nums[i-1])
for i in range(1, n + 1 ...L=[2,8,3,50]
s=sorted(L)
min=s[-1]
max=s[0]
for i in range(1,len(s)):
max=max*s[i]
lcm=min
for m in range(max,min,-1):
for i in range(len(s)):
if m%s[i]==0 and i!=len(s)-1 ...n=2
if n%2:
print (n-1)**2/4
else:
print n*(n-2)/4
n=3
print int((n ** 3 + 5 * n + 6) / 6)
# -*- coding: utf-8 -*-
N=2
b=float(N)
a=0.0
i=b
while i > 0:
a += N/i
i -= 1
print "%.2f" % a
# 已经有n张卡片，得到下一张与手上不同的卡片的概率是N-n/N，期望是N/(N-n), 1+N/(N-1)+N/(N-2)+.......+N/(1)
L = "caayyhheehhbbbhhjhhyyaac"
start, length = 0, 0
for i in range(1,len(L)-1):
if L[i] == L[i+1]:
s = min(len(L[:i+1]),len(L[i+1:]))
a, b = 1, 1
while a < s:
if L[i-a] == L[i+1+a]: # to the left 1 pace ...# find if n*(n-1)/2 can be separated into two adjacent numbers
def adjacent(num):
a=round(num**(1/2.0),0)
if a*(a+1)==num:
return True
god_num = 0
for n in range(2,N+1):
if adjacent(n*(n-1)/2):
god_num+=1
print n
print(god_num ...# -*- coding: utf-8 -*-
# s = 0
# for i in range(1,N+1):
# s = s + i**2*(N//i) # step 1
# print (s)
# N = 16
# L1 = list(range(1,N+1))
# L2 = [N//i for i in L1] # step 2 [16, 8, 5, 4, 3, 2, 2, 2, 1, 1, 1, 1 ...