No_38Pandigital multiples

最新推荐文章于 2017-06-27 10:28:00 发布

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本文探讨如何通过将整数与其序列（1到n）相乘并进行串联操作，形成一个包含所有从1到9数字的Pandigital九位数，并找出最大的可能值。

Take the number 192 and multiply it by each of 1, 2, and 3:

192  1 = 192
192  2 = 384
192  3 = 576

By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)

The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).

What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... ,n) where n 1?

import time 
t=time.time()
def CheckPandigital(num):
    """
    判断 num是不是Pandigital，如果是返回n值，否则返回1
    """
    numstr=""
    maxn=1
    while 1:
        if len(numstr)==9:
            if set([str(x) for x in range(1,10)])- set(numstr):#判断1，9个数字是否在numstr里
                return 1
            else:
                return maxn
        if len(numstr)>9:
            return 1
        else:
            numstr+=str(num*maxn)
            maxn+=1
if __name__ == '__main__':
    maxpandigital=1
    for num in xrange(98765,0,-1):
        if CheckPandigital(num)>1:
            maxpandigital=num
            break
    print maxpandigital,maxpandigital*2
    print time.time()-t

9327 18654

0.137000083923