Python：每日一题 88（答题领鱼币）
题目：列出所有3个正整数，使它们之和等于20，并算出有多少种可能排列，如
1 1 18
1 2 17
......
总共有？？种排列
注：不同的排列次序各算一次，如
1 1 18
1 18 1
18 1 1
都要列出。
我的解法
**** Hidden Message ***** 本帖最后由 jerryxjr1220 于 2017-8-31 09:27 编辑
from itertools import product
print(len(['^_^' for each in product(range(1,19), repeat=3) if sum(each)==20])) 本帖最后由 小星星LLL 于 2017-8-31 10:32 编辑
import math
# A(n,m)＝n(n-1)(n-2)……(n-m+1)＝n!/(n-m)!
# C(n,m)=A(n,m)/A(m,m)=A(n,m)/m!
print(math.factorial(19)/math.factorial(19-3))
# count = 3!*C(19,3)=19!*3!/((19-3)!*3!)=19!/(19-3)!
排列组合就好啦
计算机是数学的应用233 def sum20():
c = 0
for i in range(1,19):
for j in range(1,19):
for k in range(1,19):
if (i+j+k) == 20:
## print(i,j,k)
c += 1
print(c)
求一下不重复，怎么算。 bush牛 发表于 2017-8-31 11:01
求一下不重复，怎么算。
那就剔除重复的即可。
from itertools import product
sets = []
for each in product(range(1,19), repeat=3):
if sum(each) == 20 and sorted(list(each)) not in sets:
sets.append(sorted(list(each)))
print(len(sets))
print(sets)
33
[, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ] 啊啊啊啊啊 一个循环 三个参数 , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , jerryxjr1220 发表于 2017-8-31 11:29
那就剔除重复的即可。
33
谢谢 本帖最后由 suloman 于 2017-8-31 16:53 编辑
t = 0
for i in range(1,19):
for j in range(1,19):
for k in range(1,19):
if i+j+k == 20:
print("%d %d %d" % (i,j,k))
t+=1
print("总共有%d种"%t) def compute():
count = 0
for i in range(1,20):
for j in range(1,20):
for k in range(1,20):
if i+j+k == 20:
print i,j,k
count +=1
print ('total is :'+str(count))
total is：171 计算 3个正整数之和等于 n 的排列可能
踏实的计算，两层循环
def cal_1(n):
result = 0
for i in range(1,n-1):
for j in range(1,n-1):
if n-i-j >0:
result += 1
return result
用itertools，笛卡尔积
这个可能是最慢的
import itertools as it
def cal_2(n):
return sum(1 for i in it.product(range(1,n-1),repeat=3) if sum(i)==n)
用numpy，3维数组，
这个应该是最快的
import numpy as np
def cal_3(n):
a = np.arange(1,n-1)
return len(np.where((a +a.reshape(-1,1) +a.reshape(-1,1,1)) ==n))
n=20 的话应该就是：171种 for i in range(1, 19):
for j in range(1, 20 - i):{:5_106:}
缩小了第二层循环的范围，还避免了判断，很巧啊
我就想着直积，没想到 count = 0
for i in range(1,21):
for j in range(1,21-i):
for k in range(1,21-i-j):
if i+j+k == 20:
print(i,j,k)
count +=1
print('一共%d中组合'% count) 刚学Python的渣渣，把最渣的代码奉上。。。。。求指教优化
import itertools
def First_Method():
#"""最简单的暴力枚举"""
result = []
for i in range(1,19):
for j in range(1,19):
k = 20 - i - j
if k < 1:
continue
tmp =
result.append(tmp)
print(result)
print("有{}种排列".format(len(result)))
def Second_Method():
"""用itertools库生成列表再赛选符合条件的"""
number =
tmp = list(itertools.product(* * 3))
result =
print(result)
print("有{}种排列".format(len(result)))
if __name__ == "__main__":
First_Method()
Second_Method()
{:9_222:} count = 0
for i in range(1, 18):
for j in range(1, 18):
for k in range(1, 18):
if 20 == (i + j + k):
print(i,j,k)
count += 1
print('共有' + str(count) + '组')
66 本帖最后由 chunchun2017 于 2017-9-1 17:12 编辑
count = 0
list0 = []
for i in range(1,19):
for j in range(1,19):
if(20-i-j>0):
list0.append((i,j,20-i-j))
count+=1
print('所有可能的排列是:',list0)
print('一共有%d种可能的排列' %count)
===================
运行结果：
所有可能的排列是: [(1, 1, 18), (1, 2, 17), (1, 3, 16), (1, 4, 15), (1, 5, 14), (1, 6, 13), (1, 7, 12), (1, 8, 11), (1, 9, 10), (1, 10, 9), (1, 11, 8), (1, 12, 7), (1, 13, 6), (1, 14, 5), (1, 15, 4), (1, 16, 3), (1, 17, 2), (1, 18, 1), (2, 1, 17), (2, 2, 16), (2, 3, 15), (2, 4, 14), (2, 5, 13), (2, 6, 12), (2, 7, 11), (2, 8, 10), (2, 9, 9), (2, 10, 8), (2, 11, 7), (2, 12, 6), (2, 13, 5), (2, 14, 4), (2, 15, 3), (2, 16, 2), (2, 17, 1), (3, 1, 16), (3, 2, 15), (3, 3, 14), (3, 4, 13), (3, 5, 12), (3, 6, 11), (3, 7, 10), (3, 8, 9), (3, 9, 8), (3, 10, 7), (3, 11, 6), (3, 12, 5), (3, 13, 4), (3, 14, 3), (3, 15, 2), (3, 16, 1), (4, 1, 15), (4, 2, 14), (4, 3, 13), (4, 4, 12), (4, 5, 11), (4, 6, 10), (4, 7, 9), (4, 8, 8), (4, 9, 7), (4, 10, 6), (4, 11, 5), (4, 12, 4), (4, 13, 3), (4, 14, 2), (4, 15, 1), (5, 1, 14), (5, 2, 13), (5, 3, 12), (5, 4, 11), (5, 5, 10), (5, 6, 9), (5, 7, 8), (5, 8, 7), (5, 9, 6), (5, 10, 5), (5, 11, 4), (5, 12, 3), (5, 13, 2), (5, 14, 1), (6, 1, 13), (6, 2, 12), (6, 3, 11), (6, 4, 10), (6, 5, 9), (6, 6, 8), (6, 7, 7), (6, 8, 6), (6, 9, 5), (6, 10, 4), (6, 11, 3), (6, 12, 2), (6, 13, 1), (7, 1, 12), (7, 2, 11), (7, 3, 10), (7, 4, 9), (7, 5, 8), (7, 6, 7), (7, 7, 6), (7, 8, 5), (7, 9, 4), (7, 10, 3), (7, 11, 2), (7, 12, 1), (8, 1, 11), (8, 2, 10), (8, 3, 9), (8, 4, 8), (8, 5, 7), (8, 6, 6), (8, 7, 5), (8, 8, 4), (8, 9, 3), (8, 10, 2), (8, 11, 1), (9, 1, 10), (9, 2, 9), (9, 3, 8), (9, 4, 7), (9, 5, 6), (9, 6, 5), (9, 7, 4), (9, 8, 3), (9, 9, 2), (9, 10, 1), (10, 1, 9), (10, 2, 8), (10, 3, 7), (10, 4, 6), (10, 5, 5), (10, 6, 4), (10, 7, 3), (10, 8, 2), (10, 9, 1), (11, 1, 8), (11, 2, 7), (11, 3, 6), (11, 4, 5), (11, 5, 4), (11, 6, 3), (11, 7, 2), (11, 8, 1), (12, 1, 7), (12, 2, 6), (12, 3, 5), (12, 4, 4), (12, 5, 3), (12, 6, 2), (12, 7, 1), (13, 1, 6), (13, 2, 5), (13, 3, 4), (13, 4, 3), (13, 5, 2), (13, 6, 1), (14, 1, 5), (14, 2, 4), (14, 3, 3), (14, 4, 2), (14, 5, 1), (15, 1, 4), (15, 2, 3), (15, 3, 2), (15, 4, 1), (16, 1, 3), (16, 2, 2), (16, 3, 1), (17, 1, 2), (17, 2, 1), (18, 1, 1)]
一共有171种可能的排列 sum0 = 0
sum1 = 0
for i in range(1,19):
for j in range(1,19):
for k in range(1,19):
if i + j + k == 20:
print('%d + %d + %d = 20 ' %(i,j,k,))
sum0 += 1
if i ==j or i == k or j == k:
sum1 += 1
print('3个整数和为20的个数为：%d' %sum0)
print('其中3个整数有重复现象的个数为：%d' %sum1) 小星星LLL 发表于 2017-8-31 10:31
排列组合就好啦
计算机是数学的应用233
你的代码我没看错的话是：1~19个数字中任取3个也就是C(19，3)，因为可以调换位置，所以你乘了个3!。而根据题意要求所选三个数字和为20，不应该是1~19种任取3个，所以你的答案与题意并不符。