数字乘积

无心人

20很可能也是10的归宿 所以可以考虑4478976的分解

无心人

haskell算10 很困难了 估计要几个小时 11 可能就要以天记了

无心人

下面就要换C语言了 预先初始化个表 到1000万的数据均存储下来

无心人

当然存储10^8的也可以的

mathe

prodd n = foldr1 (*) [fromEnum c - fromEnum '0' | c<-show(n), c/='0'] count n |0<=n&&n<=9 = 0 |otherwise = (1+ count (prodd n)) smallfac n |n==1 = True |(mod n 2)==0 = smallfac (div n 2) |(mod n 3)==0 = smallfac (div n 3) |(mod n 5)==0 = smallfac (div n 5) |(mod n 7)==0 = smallfac (div n 7) |otherwise = False

mathe

*Main> [c|c<-[1..10000], count c==2, smallfac c] [25,27,28,35,36,45,48,54,56,63,64,72,84,125,126,128,135,144,162,192,216,224,225, 243,245,250,252,256,270,280,315,324,350,360,375,405,432,441,450,480,504,512,525, 540,560,567,576,588,625,630,640,675,720,735,756,840,945,1029,1125,1134,1152,1176 ,1215,1225,1250,1260,1280,1296,1323,1350,1372,1440,1458,1512,1536,1568,1620,1728 ,1920,2016,2025,2058,2160,2187,2205,2240,2250,2304,2352,2430,2450,2500,2520,2560 ,2592,2625,2700,2800,2835,2916,3024,3072,3125,3150,3240,3402,3500,3528,3600,3750 ,4032,4050,4116,4320,4375,4410,4500,4704,4800,5040,5103,5120,5145,5184,5250,5292 ,5400,5600,5625,5670,5760,5832,5880,6075,6125,6250,6300,6400,6561,6615,6750,6912 ,7056,7200,7203,7350,7560,8400,8505,8575,9216,9261,9450]

mathe

不过接下去构造也挺麻烦，比如25可以生成55,155,515,551,1155等等，当然它们大部分不合法（比如55含因子11） 看来一定要用c来构造了，haskell很难处理这么复杂的东西了

无心人

要删除中间带0的

无心人

不是， haskell最适合处理这个了 但你不会，呵呵

mathe

count==2的时候可以带0的（只要带0就肯定count==2），但是后面就都不允许了。 当然如果象你那样定义，可以少很多