比赛链接:Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) 官方题解:Technocup 2020 — Elimination Round 2 + Codeforces Round 596: analysis A. Forgetting Things 题意 有两个数 aa 和bb,给定 aa 的首位 dada 和bb 的首位 dbdb,问能否构造出 aa 和bb,满足 a+1=ba+...
比赛链接:Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) 官方题解:Technocup 2020 — Elimination Round 2 + Codeforces Round 596: analysis A. Forgetting Things 题意 有两个数 aa 和bb,给定 aa 的首位 dada 和bb 的首位 dbdb,问能否构造出 aa 和bb,满足 a+1=ba+...
Problem Writer:MikeMirzayanov Problem Writer:MikeMirzayanov My solution to Div.2 E : 1) Perform DFS starting at Vertex a. This DFS function returns when reached at Vertex b. 2) Perform DFS starting at Vertex b. This DFS function returns when reached at Vertex a. 3) When each function sav...
This weekend, at Nov/14/2021 09:05 (Moscow time) we will hold Codeforces Round 755 (Div. 1, based on Technocup 2022 Elimination Round 2) and Codeforces Round 755 (Div. 2, based on Technocup 2022 Elimination Round 2). They are based on problems of Technocup 2022 Elimination Round 2 ...
for(int now=2;now*now<=a[i];now++){ int number=0; while(a[i]%now==0){ a[i]/=now; number+=1; } if(number%k) fac.push_back(make_pair(now,number%k)); } if(a[i]>1)fac.push_back(make_pair(a[i],1%k)); vector<pair<int,int>>fac2; ...
Codeforces Round #596 (Div. 2, based on Technocup 2020 Elimination Round 2) A. Forgetting Things,链接:https://codeforces.com/contest/1247/problem/A题意:Kolyaisveryabsentminded.Todayhismathteacheraskedhimtosolveasimpleproblemwiththe
Wabbit 正在玩一个游戏,有 n 个 Boss,编号从 1 到 n。 Boss 可以按任意顺序进行战斗。 每个 Boss 都需要被击败一次。 有一个参数叫boss Bonus,初始值为0。 当第i 个 Boss 被击败时,当前的 Boss 加成会被添加到 Wabbit 的分数上,然后 boss 加成的值会增加点增量 ci。 请注意,ci 可以为负数,这意味着...
Codeforces Technocup 2017 - Elimination Round 1 题解 只看楼主 收藏 回复quailty 算法狂人 11 RT quailty 算法狂人 11 A. 倒着做,可以发现如果b是偶数只能进行操作1,否则只能进行操作2B. 抠出所有数字直接加起来C. 询问a[1]+a[2]、a[2]+a[3]和a[3]+a[1]的值之后可以解出a[1]、a[2]...
Codeforces Round #606 (Div. 2, based on Technocup 2020 Elimination Round 4) 2019-12-15 12:46 − ##A - Happy Birthday, Polycarp! 题意:给一个n m; void test_case() { int n, ans = 0; scanf("%d", &n); for(int i = 1; i > tmp] = max(m[a[i] >> tmp], tmp... ...
1#include<bits/stdc++.h>2using namespace std;3typedef long long ll;4llgcd(ll a,ll b)5{6returnb==0?a:gcd(b,a%b);7}8intmain(void)9{10ll n,k;11cin>>n>>k;12ll temp=1;13for(ll i=1;i<=k;i++)14temp*=10;15cout<<n*temp/gcd(temp,n);16}...