51nod1245

题意

51nod

做法

(p)在({mchoose k})中出现的次幂为：
(sumlimits_{i=1}^{infty} leftlfloorfrac{m}{p^i} ight floor-sumlimits_{i=1}^{infty} leftlfloorfrac{k}{p^i} ight floor-sumlimits_{i=1}^{infty} leftlfloorfrac{m-k}{p^i} ight floor)

然后单独考虑(leftlfloorfrac{m}{p^i} ight floor-leftlfloorfrac{k}{p^i} ight floor-leftlfloorfrac{m-k}{p^i} ight floor)，([leftlfloorfrac{m}{p^i} ight floor-leftlfloorfrac{k}{p^i} ight floor-leftlfloorfrac{m-k}{p^i} ight floor=1]Longrightarrow m\%p^i<k\%p^i)

(p)在({mchoose k})中出现的次幂为：在(p)进制下，(k)某后缀大于(m)某后缀的次数
数位dp乱搞就好了