赛后听 Forever97 讲的思路,强的一匹- - /* CodeForces 839D - Winter is here [ 数论,容斥 ] | Codeforces Round #428 (Div. 2) 题意: 给出数列a[N] 对每个子集,若 gcd(a[I1], a[I2], a[I3] ..., a[In]) > 1,则贡献为 n*gcd 求总贡献和 限制: N <= 2e5,a[i] <= 1e6 分析: 记录 num[i]数组为 i 的倍数的个数 则 gcd >=…

起初误以为到每个叶子的概率一样于是.... /* CodeForces 839C - Journey [ DFS,期望 ] | Codeforces Round #428 (Div. 2) */ #include <bits/stdc++.h> using namespace std; const int N = 100005; int n; vector<int> G[N]; double dp[N], val[N]; bool vis[N]; void dfs(int u, i…

血崩- - /* CodeForces 839B - Game of the Rows [ 贪心,分类讨论] | Codeforces Round #428 (Div. 2) 注意 2 7 2 2 2 2 2 2 2 这组- - */ #include <bits/stdc++.h> using namespace std; int n, n2, n4, a[105], k; bool solve() { n2 = 2*k; n4 = k; for (int i = 1; i <= n;…

D. Winter is here 题目连接: http://codeforces.com/contest/839/problem/D Description Winter is here at the North and the White Walkers are close. John Snow has an army consisting of n soldiers. While the rest of the world is fighting for the Iron Throne,…

给你一个序列,让你对于所有gcd不为1的子序列,计算它们的gcd*其元素个数之和. 设sum(i)为i的倍数的数的个数,可以通过容斥算出来. 具体看这个吧:http://blog.csdn.net/jaihk662/article/details/77161436. 注意1*C(n,1)+2*C(n,2)+...+n*C(n,n)=n*2^(n-1). #include<cstdio> using namespace std; typedef long long ll; #define MOD…

Winter is here at the North and the White Walkers are close. John Snow has an army consisting of n soldiers. While the rest of the world is fighting for the Iron Throne, he is going to get ready for the attack of the White Walkers. He has created a m…

题目链接:http://codeforces.com/contest/839 A. Arya and Bran 题意:每天给你一点糖果,如果大于8个,就只能给8个,剩下的可以存起来,小于8个就可以全部给完 解法:水题,按照题意模拟一下即可. #include <bits/stdc++.h> using namespace std; int n, k, a[110]; int main() { cin>>n>>k; int s = 0; for(int i=1; i<…

D. Winter is here time limit per test:3 seconds memory limit per test:256 megabytes input:standard input output:standard output Winter is here at the North and the White Walkers are close. John Snow has an army consisting of n soldiers. While the res…

[题目链接] http://codeforces.com/contest/839/problem/D [题目大意] 给出一些数,求取出一些数,当他们的GCD大于0时,将数量乘GCD累加到答案上, 求累加和. [题解] 我们枚举GCD,统计为其倍数的数字数量,先假定其能组成的集合数为贡献, 但是我们发现在统计的过程中有多余统计的部分,比如4和8是2的倍数, 然而它们的GCD等于4,所以我们对于每个集合数的贡献要减去所有其倍数的集合数的贡献, 倒着容斥即可. [代码] #include <cstdi…

http://codeforces.com/contest/839/problem/E 最大团裸题= =,用Bron–Kerbosch算法,复杂度大多博客上没有,维基上查了查大约是O(3n/3) 最大团: V中取K个顶点,两点间相互连接 最大独立集: V中取K个顶点,两点间不连接 最大团数量 = 补图中最大独立集数 //#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast,no-st…