首页 > > 网络编程 > 其它 >

python计算1~2008中0和1的个数

2018-06-18 03:26:11来源：未知阅读 ()

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)