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python计算1~2008中0和1的个数

时间:2018-06-11 13:44:17      阅读:254      评论:0      收藏:0      [点我收藏+]

标签:pen   字符   list   IV   print   odi   python   col   nbsp   

计算1~2008中所有自然数中1和0的个数总数。
  • 通过自然数的大小划分区间,将自然数每位上的数载入列表,循环计数。
list = []
onecount = 0
zerocount = 0
for i in range(1,2009):
     if int(i/1000)!=0:                           #thousands
         list.append(int(i/1000))
         list.append(int(i/100)%10)
         list.append(int(i/10)%10)
         list.append(i%10)
         for count in list:
             if count == 1:
                 onecount = onecount + 1
             if count == 0:
                 zerocount = zerocount + 1
         list=[]                                    #clear
     elif int(i/100)!=0:                            #hundreds
         list.append(int(i / 100))
         list.append(int(i / 10)%10)
         list.append(i % 10)
         for count in list:
             if count == 1:
                 onecount = onecount + 1
             if count == 0:
                 zerocount = zerocount + 1
         list = []                                    #clear
     elif int(i/10)!=0:                              #tens
         list.append(int(i / 10))
         list.append(i%10)
         for count in list:
             if count == 1:
                 onecount = onecount + 1
             if count == 0:
                 zerocount = zerocount + 1
         list = []                                  #clear
     else:                                         #single figurea
         if i == 1:
             onecount = onecount + 1
         if i == 0:
             zerocount = zerocount + 1
print(1的个数为:,onecount)
print(0的个数为:,zerocount)
  • 将数转为字符串,循环统计字符串中‘0’,‘1’的个数。
#coding:GBK
#计算0的个数和1的个数
s0=0
for i in range(1, 2009):
    c=str(i)
    s0=s0+c.count(0)
print(1-2008中0的个数为:)
print(s0)
s1=0
for i in range(1, 2009):
    c=str(i)
    s1=s1+c.count(1)
print(1-2008中1的个数为:)
print (s1)

 

python计算1~2008中0和1的个数

标签:pen   字符   list   IV   print   odi   python   col   nbsp   

原文地址:https://www.cnblogs.com/FerrisFu/p/9166490.html

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