摘要: LL fenkuai(LL k){ LL j; LL ans=0; for (LL i=1;i<=k;i=j+1){ j=n/(n/i); ans+=(j-i+1)%mod*((n/i)%mod)%mod; ans%=mod; } return ans; } 阅读全文
posted @ 2020-08-02 10:04 xlinsist 阅读(67) 评论(0) 推荐(0)