比赛链接:https://atcoder.jp/contests/abc170 A - Five Variables 题意 $5$ 个数中有 $1$ 个 $0$,判断是第几个. 代码 #include <bits/stdc++.h> using namespace std; int main() { for (int i = 1; i <= 5; i++) { int x; cin >> x; if (x == 0) { cout << i << &q…
D - Disjoint Set of Common Divisors Problem Statement Given are positive integers AA and BB. Let us choose some number of positive common divisors of AA and BB. Here, any two of the chosen divisors must be coprime. At most, how many divisors can we c…
题意:有一长度为\(n\)的数组,求该数组中有多少元素不能整除其它任一元素的个数. 题解:刚开始写了个分解质因数(我是傻逼),后来发现直接暴力枚举因子即可,注意某个元素出现多次时肯定不满足情况,再特判数组中存在\(1\)的情况即可. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include…
AtCoder Beginner Contest 137 F 数论鬼题(虽然不算特别数论) 希望你在浏览这篇题解前已经知道了费马小定理 利用用费马小定理构造函数\(g(x)=(x-i)^{P-1}\) \[x=i,g(x)=0\] \[x\ne i ,g(x)=1\] 则我们可以构造 \[f(x)=\sum^{i=0}_{P-1}(-a_i*(x-i)^{P-1}+a_i)\] 对于第\(i\)条式子当且仅当\(a_i=1 \ and \ x=i\)时取到\(1\) 代码写的比较奇怪 const…
A - ABCxxx Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement This contest, AtCoder Beginner Contest, is abbreviated as ABC. When we refer to a specific round of ABC, a three-digit number is appended after ABC. For example,…
AtCoder Beginner Contest 238 \(A - F\) 题解 A - Exponential or Quadratic 题意 判断 \(2^n > n^2\)是否成立? Solution 当 \(n\) 为 2,3,4 的时候不成立,否则成立 Code #include <bits/stdc++.h> using namespace std; using LL = long long; int main() { int n; cin >> n; bool…
A - ABC/ARC Time limit : 2sec / Memory limit : 256MB Score : 100 points Problem Statement Smeke has decided to participate in AtCoder Beginner Contest (ABC) if his current rating is less than 1200, and participate in AtCoder Regular Contest (ARC) oth…
KYOCERA Programming Contest 2021(AtCoder Beginner Contest 200) 题解 哦淦我已经菜到被ABC吊打了. A - Century 首先把当前年份减去\(1\),对应的世纪也减去\(1\),然后我们就发现第\(0\)到\(99\)年对应第\(0\)世纪,第\(100\)到\(199\)年对应第\(1\)世纪,以此类推. 答案就是\(\lfloor \frac {N-1} {100} \rfloor\).这里\(\lfloor x \rflo…
AtCoder Beginner Contest 136 题目链接 A - +-x 直接取\(max\)即可. Code #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 2e5 + 5; int main() { ios::sync_with_stdio(false); cin.tie(0); int a, b; cin >> a >> b; cout &…
AtCoder Beginner Contest 075 C bridge 桥就是指图中这样的边,删除它以后整个图不连通.本题就是求桥个数的裸题. dfn[u]指在dfs中搜索到u节点的次序值,low[u]指dfs栈中u能追溯到的最早栈中节点从次序号.这里写的很好. #include<bits/stdc++.h> using namespace std; ; vector<int> G[maxn]; int dfn[maxn],low[maxn]; int n,m,u,v,dep,…