We will hold AtCoder Regular Contest 186. Contest URL:https://atcoder.jp/contests/arc186 Start Time:http://www.timeanddate.com/worldclock/fixedtime.html?iso=20241027T2100&p1=248 Duration: 120 minutes Number of Tasks: 5 Writer:nuip Tester:maroonrk Rated range: 1200 ~ 2799 The point value...
#include<vector>#include<iostream>#include<algorithm>usingi64 =longlong;usingvint = std::vector<int>;usingPII = std::pair<int,int>;intmain(void){//Think twice, code once.std::ios::sync_with_stdio(false); i64 a[3]; std::cin >> a[0] >> a[1] >> a[2]; i64 x = a[0] ...
LL pre[N], suf[N], b[N]; voidsolve(){ cin >> n; for(inti =1; i <= n; i++) { cin >> a[i]; b[i] = a[i]; l[i] = i -1; r[i] = i +1; pre[i] = pre[i -1] + a[i]; } for(inti = n; i >=1; i--) { suf[i] = suf[i +1] + a[i]; } w...
枚举左下角矩形从哪个位置进入,组合计数一下就好了。 1#include<bits/stdc++.h>2#defineN (200009)3#defineLL long long4#defineMOD (1000000007)5usingnamespacestd;67LL inv[N],fac[N],facinv[N];8LL h,w,a,b;910voidInit()11{12fac[0]=facinv[0]=inv[1]=1;13for(inti=1; i<=200000; ...
AtCoder Beginner Contest 399(C-F详细题解) C 思路:作为一个简单无向图,有 n 个点只能有 n-1 条边,那我们只需要统计不同连通图上的点计算出所需的边即可,最后就是总边数减去所需的边即可。 代码:vector<vector<int>>… 秋日薄雾 AtCoder Beginner Contest 401 A-G 简易题解,如果题解...
164 -- 1:43:11 App [6/8] AtCoder Beginner Contest 286 (unrated) 181 -- 27:25 App [6/6] VP AtCoder Beginner Contest 166 (随机 VP) 441 1 2:00:59 App [5/7] Educational Codeforces Round 142 (1660 ⭜ 1889) 468 -- 42:43 App AtCoder Beginner Contest 334 A 至 G 題...
带入上式①,得Xmin=max(0,⌊B−1k⌋+1−A);③ 式③解释: 首先它是非负整数,所以它不能小于零, 其次用X表示Y后的式子也不能小于零,把 Y=0 代入式子①,反算X得到另一个值。 再考虑X+Y。 由③+② 得(X+Y)_{min}=(k+1)max(0, ⌊\frac{B−1}{k} ⌋+1−A))+kA+B④ ...
We will hold AtCoder Regular Contest 151. Contest URL:https://atcoder.jp/contests/arc151 Start Time:http://www.timeanddate.com/worldclock/fixedtime.html?iso=20221016T2100&p1=248 Duration: 120 minutes Number of Tasks: 6 Writer:leaf1415 ...
AtCoder Regular Contest 102D - All Your Paths are Different Lengths,题解:我们不妨总是令n=20n=20n=20。对于2<=i<=192<=i<=192iii与i+1i+1i+1之间连两条边,一条权值为000,一条权值为220−i−1220−i−12^{20}−i−1。这样
AtCoder Regular Contest 144(简称ARC 144)是AtCoder举办的一场定期比赛,旨在提供一个平台供程序员们展示他们的编程技能和解决问题的能力。本文将详细解析ARC 144的题目,并提供相应的解题思路和方法。 正文内容: 1.第一大点:题目一解析 1.1小点:题目背景介绍 在这里,我们将介绍题目的背景信息,包括题目的出处和相关...