把库文件夹直接拖到你的编译器库文件相同目录下。Mingw 的路径应该都是\lib\gcc\x86_64-w64-mingw32\8.1.0\include\c++,如果不是也容易自己寻找一下。或者定义环境变量CPLUS_INCLUDE_PATH,然后把ac-library文件夹的路径粘进去。 需要使用时直接引用库#include <atcoder/all>,并using namespace atcoder;。
AC Library Document(ja) The documents of master branch are as follows: AC Library Document(en, master) AC Library Document(ja, master) Policy Our goal of this project is to achieve that Enable every AtCoder users to use this library with minimum efforts of studying about PC Maximize convenie...
这个库是我在做atcoder的时候看到的,是人家oj自带的一个库,实际用处不大,但是可以塞到自己的板子里面,打线上的比赛用。(代码太长了,我放到的文章最后 首先是单点修改区间查询的线段树( 区间修改的搞出来了,看完这一篇的可以去看看qwq 严格鸽:像使用stl一样使用线段树——区间修改 与 实例46 赞同 · 6 评论...
every AtCoder users can use this library with minimum efforts of studying about PC maximize convenience for the usage in competitive programming. We completely ignore other usages. no bugs. This is a fantasy, but we pursue this. By this policy, we ignore some manners of C++ intentionally. For...
ModInt是Atcoder Library中的一个重要组件,它是用于处理模运算的工具。在计算机科学领域,模运算指的是将结果限制在某个特定范围内的运算,常用于解决数论问题和对大数进行高效计算。ModInt提供了一种方便和高效的方式来执行模运算。 2.2 Atcoder Library简介: Atcoder Library是由日本一个在线编程竞赛平台Atcoder推出的...
Mint 是实现的自动取模类,在 atcoder 上,也可以采用 ac-library 中的 modint。 void solve() { int h, p; cin >> h >> p; Mint dp[h+1]; dp[0] = 0; dp[1] = 1; for(int i = 2; i <= h; i++) { dp[i] = 1 + (dp[i-2] * p / 100 + dp[i-1] * (100-p) /...
To create a document that provides an introduction and guidance on utilizing the AtCoder problem library.The document should be written in a bilingual format, with one paragraph in English and the corresponding paragraph in Chinese.The writing style and grammar structure should vary randomly to avoid...
线段树,单点更新,区间查询。直接用AC-Library模板即可。 时间复杂度\(\mathcal{O}((N+Q)\log N)\)。 #include <bits/stdc++.h> #include <atcoder/segtree> using namespace std; int op(int a, int b) { return a ^ b; } int e() { return 0; } ...
In most programming languages, you can sort array of string in lexicographical order by means of sorting function standard library (e.g. std::sort in C++). When reordering the restaurants in accordance with the problem statement, you can implement it easily if you use "pair" type in C++. ...
C代码 #include<bits/stdc++.h>usingnamespacestd; /*子串 (substring) vs 子序列 (subsequence) 子串是原序列中的连续的一段。比如对于 pekinguniversity,其子串可以是 peking,可以是 univ,可以 gun,可以是空(空串也是子串),也可以是整个字符串。子串的数量是 $n(n+1)/2$。