摘要: calculate the f(n) . (3<=n<=1000000)f(n)= Gcd(3)+Gcd(4)+…+Gcd(i)+…+Gcd(n).Gcd(n)=gcd(C[n][1],C[n][2],……,C[n][n-1])C[n][k] means the number of way to c 阅读全文
posted @ 2016-11-25 12:21 lzw4896s 阅读(160) 评论(0) 推荐(0) 编辑