摘要: A 分块, 签到 int main() { IOS; int cas = 0; for (cin >> _; _; --_) { cin >> n; __int128 cur = 0; for (int i = 1, j; i <= n; i = j + 1) { j = min(n, n / (n 阅读全文
posted @ 2020-10-28 18:18 洛绫璃 阅读(183) 评论(0) 推荐(0) 编辑