第i位的总花费就是∑j=1N−ii×10j×ai相当于给高精度的数0∼N−i位加上i×ai 差分处理,再处理一下进位即可 点击查看代码 #include<bits/stdc++.h>usingnamespacestd;intn;longlongans[300001],c[200001];chars[200001];boolst;intmain(){scanf("%d%s",&n,s +1);for(inti =1;i <= n;i ...
AtCoder Beginner Contest 379 A - Cyclic#题意#输入3个连续字符a,b,c,输出另外两种顺序。思路#模拟。 代码#点击查看代码 #include<bits/stdc++.h> using namespace std; #define int long long typedef pair<int, int> pii; const int mxn =
Toyota Programming Contest 2024#11(AtCoder Beginner Contest 379) - AtCoderatcoder.jp/contests/abc379 B 直接寻找连续 k 个牙的个数即可,遍历一遍即可。 #include<bits/stdc++.h>usingnamespacestd;#define cout std::cout#define int long long#define uint unsigned long long#define double long doub...
Atcoder Education DP Contest 很好的 Atcoder 的 DP 题单,写一份题解记录一下。 Tasks - Educational DP ContestA Frog 1题意从 1 号点跳到 n 号点,每次可以从 i 跳到 i+1 或者 i+2,若从 i 跳到 j ,其花费为:|h_j-h… Jackl...发表于好题集锦 AtCoder Beginner Contest 359 (A-G个人题...
AtCoder Beginner Contest 359 A~F https://www.bilibili.com/video/BV1G4gveBENB/ AtCoder Beginner Contest 359 | 完成度 [6 / 7] https://www.bilibili.com/video/BV1kS411A7M3/ AtCoder Beginner Contest 359 A 至 G 題讲解 by dreamoon https://www.bilibili.com/video/BV1x1421r7gH/ AtCoder...
AtCoder Beginner Contest 353A-E+G https://www.bilibili.com/video/BV1wt421M7Z5/ https://www.bilibili.com/video/BV1TZ421774X/ AtCoder Beginner Contest 353 CDEG讲解 https://www.bilibili.com/video/BV1rm421u7cs/ AtCoder Beginner Contest 353, f题写懵了 https://www.bilibili.com/video/...
We will hold AtCoder Beginner Contest 137. Contest URL: https://atcoder.jp/contests/abc137 Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20190810T2100&p1=248 Duration: TBD (around 2 hours) Number of Tasks: 6 writer:potetisensei HIR180 Drafear yokozuna57 DEGwer ...
AtCoder Beginner Contest 180 个人题解(快乐DP场) Here A - box 输出\(N - A + B\) B - Various distances 按题意输出 3 种距离即可 #include <bits/stdc++.h> using namespace std; using ll = long long; int main() { ios_base::sync_with_stdio(false), cin.tie(0);...
AtCoder Beginner Contest 193 部分题解 E - Oversleeping 求是否存在\(t\)满足\(t=t_1(mod (2X+2Y)) and t=t_2(mod (P+Q))\) 注意到\(Q\)和\(Y\)非常小,直接枚举套个\(exCRT\)就行了(虽然赛场上没看出来,\(exCRT\)也忘了记得快速乘...
We will hold AtCoder Beginner Contest 167. Contest URL: https://atcoder.jp/contests/abc167 Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20200510T2100&p1=248 Duration: 100 minutes Number of Tasks: 6 Writer: gazelle, kort0n, kyopro_friends, potetisensei, she...