Numerical Sequence (easy version)

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Numerical Sequence (easy version)

http://codeforces.com/problemset/problem/1216/E1 E1. Numerical Sequence (easy version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output The only difference between the easy and the hard ver…

cf1216E2 Numerical Sequence (hard version)（思维）

cf1216E2 Numerical Sequence (hard version) 题目大意一个无限长的数字序列,其组成为\(1 1 2 1 2 3 1.......1 2 ... n...\),即重复的\(1~1,1~2....1~n\),给你一个k,求第k(k<=1e18)个数字是什么 solution 暴力枚举妥妥T掉手摸能看出来每一个单独的序列是等差数列然后每一个序列的数字个数也是等差数列如果预处理出等差数列的数组,那么询问的时候直接调用即可参考lzz的blogs 说说这一段…

[CF1216E] Numerical Sequence hard version

题目 The only difference between the easy and the hard versions is the maximum value of k. You are given an infinite sequence of form "112123123412345…" which consist of blocks of all consecutive positive integers written one after another. The fi…

cf1216E2 Numerical Sequence (hard version) 二分查找、思维题

题目描述 The only difference between the easy and the hard versions is the maximum value of k. You are given an infinite sequence of form "112123123412345--" which consist of blocks of all consecutive positive integers written one after another. The…

【二分】CF Round #587 (Div. 3)E2 Numerical Sequence (hard version)

题目大意有一个无限长的数字序列,其组成为1 1 2 1 2 3 1.......1 2 ... n...,即重复的1~1,1~2....1~n,给你一个\(k\),求第\(k(k<=10^{18})\)个数字是什么. 思路设\(a[i]\)为该数到达\(i\)的长度,\(b[i]\)为第\(i\)个数那一块的长度,\(\sum b[i]=a[i]\),发现\([10^i,10^{i+1})\)的数长度均为\(i+1\),那么在这一段中\(b[i]\)都是一个等差数列,而计算\(a[i]\)就…

CF1264D1 Beautiful Bracket Sequence (easy version)

考虑在一个确定的括号序列中,我们可以枚举中间位置,按左右最长延伸出去的答案计算. 我们很自然的思考,我们直接维护左右两边,在删除一些字符后能够延伸的最长长度. 我们设\(f_{i,j}\)为\(i\)点合法删除向左延伸的最大长度. \( f_{i,j} = \left\{ \begin{aligned} &f_{i - 1,j} (a[i] = ')'\ )\\ &f_{i - 1,j - 1}(a[i] = ')'\ )\\ &f_{i - 1,j} + f_{i - 1 ,j…

Ping-Pong (Easy Version)（DFS）

B. Ping-Pong (Easy Version) time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output In this problem at each moment you have a set of intervals. You can move from interval (a, b) from our set to in…