Project Euler：Problem 93 Arithmetic expressions

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的所有组合情况，保存为list

对于每一种组合都有24种排列情况

每一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，而且每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。

最终是要比较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]]+resultdef 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 numdef 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/bcomb=xcombination([i for i in range(10)],4)comb_list=list(comb)bestprem=[0 for i in range(4)]bestres=0for 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)