题目:有四个数字:1、2、3、4,能组成多少个互不相同且无重复数字的三位数?各是多少?
程序分析:可填在百位、十位、个位的数字都是1、2、3、4。组成所有的排列后再去 掉不满足条件的排列。
程序源代码:
以上实例输出结果为:
1 2 3 1 2 4 1 3 2 1 3 4 1 4 2 1 4 3 2 1 3 2 1 4 2 3 1 2 3 4 2 4 1 2 4 3 3 1 2 3 1 4 3 2 1 3 2 4 3 4 1 3 4 2 4 1 2 4 1 3 4 2 1 4 2 3 4 3 1 4 3 2
zavier
126***[email protected]
使用列表形式,并计算总结:
#!/usr/bin/python # -*- coding: UTF-8 -*- # 原答案没有指出三位数的数量,添加无重复三位数的数量 d=[] for a in range(1,5): for b in range(1,5): for c in range(1,5): if (a!=b) and (a!=c) and (c!=b): d.append([a,b,c]) print "总数量:", len(d) print d
zavier
126***[email protected]
盼盼
946***[email protected]
将for循环和if语句综合成一句,直接打印出结果
#!/usr/bin/env python # -*- coding: UTF-8 -*- list_num = [1,2,3,4] list = [i*100 + j*10 + k for i in list_num for j in list_num for k in list_num if (j != i and k != j and k != i)] print (list)
盼盼
946***[email protected]
习惯乌龙茶
rea***[email protected]
参考方法(设置最大,最小值):
#!/usr/bin/python
# -*- coding: UTF-8 -*-
line=[]
for i in range(123,433):
a=i%10
b=(i%100)//10
c=(i%1000)//100
if a!=b and b!=c and a!=c and 0<a<5 and 0<b<5 and 0<c<5 :
print (i)
line.append(i)
print('the total is :',len(line))
习惯乌龙茶
rea***[email protected]
python3 下参考方案:
#!/usr/bin/env python3
#coding:utf-8
num=[1,2,3,4]
i=0
for a in num:
for b in num:
for c in num:
if (a!=b) and (b!=c) and (c!=a):
i+=1
print(a,b,c)
print('总数是:',i)
白色帽子
liu***[email protected]
参考方法:
#!/usr/bin/env python
#-*- coding:utf-8 -*-
#用集合去除重复元素
import pprint
list_num=['1','2','3','4']
list_result=[]
for i in list_num:
for j in list_num:
for k in list_num:
if len(set(i+j+k))==3:
list_result+=[int(i+j+k)]
print("能组成%d个互不相同且无重复数字的三位数: "%len(list_result))
pprint.pprint(list_result)
白色帽子
liu***[email protected]
Chyroc
che***[email protected]
python自带这个函数的
#!/usr/bin/env python3 #coding:utf-8 from itertools import permutations for i in permutations([1, 2, 3, 4], 3): print(i)
Chyroc
che***[email protected]
weapon
965***[email protected]
补充一下:
#!/usr/bin/env python3 # -*- coding:utf-8 -*- #补充一下 from itertools import permutations for i in permutations([1, 2, 3, 4], 3): k = '' for j in range(0, len(i)): k = k + str(i[j]) print (int(k))
weapon
965***[email protected]
逸章
me@***gdewen.com
没事找事之位运算
# coding:utf-8 #从 00 01 10 到 11 10 01 for num in range(6,58): a = num >> 4 & 3 b = num >> 2 & 3 c = num & 3 if( (a^b) and (b^c) and (c^a) ): print a+1,b+1,c+1
逸章
me@***gdewen.com
DCGDDD
805***[email protected]
考虑减少冗余判断和循环,做如下优化;
#!/usr/bin/python3 for i in range(1, 5): for j in range(1, 5): if (j==i) : continue; for k in range(1, 5): if (k==i or k==j): continue; print(i,j,k);
DCGDDD
805***[email protected]
嘿嘿
123***.com
Python3 测试实例:
#!/usr/bin/python3 list = [1,2,3,4] for i in list: list1 = list.copy() list1.remove(i) for j in list1: list2 = list1.copy() list2.remove(j) for k in list2: print(i, j, k)
嘿嘿
123***.com
Krystal
104***[email protected]
加入了format函数
#!/usr/bin/python
#-*- coding: UTF-8 -*-
list_num = [1,2,3,4]
list = [i*100 + j*10 + k for i in list_num for j in list_num for k in list_num if ( i != j and i != k and j != k)]
d = len(list)
print('1,2,3,4能组成 %d 个互不相同且无重复数字的三位数。' % d)
print('他们各是:%s' % list)Krystal
104***[email protected]
红烧土豆块
545***[email protected]
数量统计用个做自加就够了
#!/usr/bin/env python3
#coding=utf-8
from itertools import permutations
t = 0
for i in permutations('1234',3):
print(''.join(i))
t += 1
print("不重复的数量有:%s"%t)红烧土豆块
545***[email protected]
这个好好玩
303***[email protected]
参考:
#encoding=utf8 #有四个数字:1、2、3、4,能组成多少个互不相同且无重复数字的三位数?各是多少? d = [] for j in range(1,5): for k in range(1,5): for l in range(1,5): if l!=j!=k!=l: d.append(int(str(j)+str(k)+str(l))) print d print len(d)
这个好好玩
303***[email protected]
XMDERO
124***[email protected]
#直接用列表推导式 [(x,y,z) for x in range(1,5) for y in range(1,5) for z in range(1,5) if(x!=y)and(x!=z)and(y!=z)]
XMDERO
124***[email protected]
阳光不锈
173***[email protected]
参考方法:
#coding=utf-8
print("----------递归法 -------------")
#递归法
def f01(i):
if i==123:
print(i)
return
else:
if (set('567890') & set(str(i))==set()) and (len(set(str(i)))==3):
print(i)
f01(i-1)
f01(432)
print("----------生成器法-------------")
#生成器法
def f02():
for i in range(123,433):
if (set('567890') & set(str(i))==set()) and (len(set(str(i)))==3):
yield i
for i in f02():
print(i)阳光不锈
173***[email protected]
whitestonex
851***[email protected]
参考方法:
import itertools
DataIn = list('1234')
TmpList = []
for x in list(itertools.combinations(DataIn,3)):
TmpList = TmpList + list(itertools.permutations(x,3))
for i in TmpList:
print(''.join(i))whitestonex
851***[email protected]