摘要: num = sqrt(n); for (int i = 1; i <= num; i++) st[i] = n / num * (i - 1) + 1, ed[i] = n / num * i; ed[num] = n; for (int i = 1; i <= num; i++) { for (i 阅读全文
posted @ 2021-11-02 21:13 _LH2000 阅读(60) 评论(0) 推荐(1) 编辑