18/04/01 15:47:58.83 NSJxVWjO.net
[n/5]+2[n/5^2]+3[n/5^3]+...+m[n/5^m]
を評価しなければいけなかった（。。）
mを非負整数とするときm[a]<=[am]
だから
[n/5]+2[n/5^2]+3[n/5^3]+...+m[n/5^m]<=
[n/5]+[2n/5^2]+[3n/5^3]+...+[mn/5^m]<=
[n/5+2n/5^2+...+mn/5^m+...]<=
[n (1/5)(1+2(1/5)+3(1/5)^2+....)]
=[n(1/5)/(1-1/5)^2]
=[5n/16]<=[n/2]
あとは同じ 👀
Rock54: Caution(BBR-MD5:0be15ced7fbdb9fdb4d0ce1929c1b82f)