题意
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