By using each of the digits from the set, {1, 2, 3, 4}, exactly once, and making use of the four arithmetic operations (+, −, *, /) and brackets/parentheses, it is possible to form
different positive integer targets.
For example,
8 = (4 * (1 + 3)) / 2
14 = 4 * (3 + 1 / 2)
19 = 4 * (2 + 3) − 1
36 = 3 * 4 * (2 + 1)
Note that concatenations of the digits, like 12 + 34, are not allowed.
Using the set, {1, 2, 3, 4}, it is possible to obtain thirty-one different target numbers of which 36 is the maximum, and each of the numbers 1 to 28 can be obtained before encountering
the first non-expressible number.
Find the set of four distinct digits, a < b < c < d, for which the longest set of consecutive positive integers, 1 to n, can be obtained,
giving your answer as a string: abcd.
先求出10选4的全部组合情况，保存为list
对于每一种组合都有24种排列情况
每个排列情况其运算顺序都有5种
关于四个数的运算涉及到3个操作符。并且每一个操作符理论上有四种选择：加减乘除。并将得出的整数运算结果标记出来。
终于是要比較每一种组合的标记出来的结果,从1到n都有标记的最大的那个n
def xcombination(seq,length):
if not length:
yield []
else:
for i in range(len(seq)):
for result in xcombination(seq[i+1:],length-1):
yield [seq[i]]+result def nextPermutation(self, num):
if len(num) < 2:
return num
partition = -1
for i in range(len(num) - 2, -1, -1):
if num[i] < num[i + 1]:
partition = i
break
if partition == -1:
return num[::-1]
for i in range(len(num) - 1, partition, -1):
if num[i] > num[partition]:
num[i], num[partition] = num[partition], num[i]
break
num[partition + 1:] = num[partition + 1:][::-1]
return num def ope(a,b,num):
if a==None or b==None:
return None
if num == 1:
return a+b
if num == 2:
return a-b
if num == 3:
return a*b
if num == 4:
if b == 0:
return None
else:
return a/b comb=xcombination([i for i in range(10)],4)
comb_list=list(comb)
bestprem=[0 for i in range(4)]
bestres=0
for prem in comb_list:
tmp=prem
flag=1
num_list=[0]*(9*8*7*6)
while tmp != prem or flag==1:
flag=0
for i in range(1,5):
for j in range(1,5):
for k in range(1,5):
num=ope(ope(ope(prem[0],prem[1],i),prem[2],j),prem[3],k)
if num!=None and num==int(num) and num > 0 and num < len(num_list):
num_list[int(num)]=True num=ope(ope(prem[0],ope(prem[1],prem[2],j),i),prem[3],k)
if num!=None and num==int(num) and num > 0 and num < len(num_list):
num_list[int(num)]=True num=ope(prem[0],ope(ope(prem[1],prem[2],j),prem[3],k),i)
if num!=None and num==int(num) and num > 0 and num < len(num_list):
num_list[int(num)]=True num=ope(prem[0],ope(prem[1],ope(prem[2],prem[3],k),j),i)
if num!=None and num==int(num) and num > 0 and num < len(num_list):
num_list[int(num)]=True num=ope(ope(prem[0],prem[1],i),ope(prem[2],prem[3],k),j)
if num!=None and num==int(num) and num > 0 and num < len(num_list):
num_list[int(num)]=True
count=1
while num_list[count]==True:
count=count+1 if count > bestres:
bestres=count
bestprem=prem prem=nextPermutation((),[prem[i] for i in range(4)]) print(bestres,' ',bestprem)