D - High Load 因为要出口节点距离最小,所以除了根节点(根节点连接k个儿子)其他节点的儿子只能有一个,其他情况下的距离都比这个长,因为如果不是这样,那么根节点连接的子树数量就小与k,那么每个子树的长度就更长了(显然,画图可证). n-1个节点平均分配到k条子树上,如果画图的话,是这样的 只要讨论一下最后一层的的个数((n-1)%k)情况即可,连接方式也如上图. 代码: #include<bits/stdc++.h> using namespace std; int main() {…

题目描述: High Load time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output Arkady needs your help again! This time he decided to build his own high-speed Internet exchange point. It should consist of…

/*题意:给你一个序列经过最小变换,变换一次为一个数+1,一个数-1,使得最大值与最小值相差1:思路:与最后得到的序列相差的sum/2:*/#include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring>…

看到这题是我的想法就是肯定跟平均值有关但是接下来就不知道怎么做了 看完大神的正解数之后,原来发现是这么简单,但是就是不知道为啥一定是平均值和平均值加1,而不是平均值和平均值减1: 好啦下面就贴出大神的代码吧: #include<iostream> #include<cstdio> #include<algorithm> using namespace std; int a[100005]; int main() { int n;scanf("%d",…

C. Load Balancing 题目连接: http://www.codeforces.com/contest/609/problem/C Description In the school computer room there are n servers which are responsible for processing several computing tasks. You know the number of scheduled tasks for each server:…

D. High Load 题目连接: http://codeforces.com/contest/828/problem/D Description Arkady needs your help again! This time he decided to build his own high-speed Internet exchange point. It should consist of n nodes connected with minimum possible number of…

C. Load Balancing time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output In the school computer room there are n servers which are responsible for processing several computing tasks. You know the…

C. Load Balancing time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output In the school computer room there are n servers which are responsible for processing several computing tasks. You know the…

847H - Load Testing 思路:dp. 代码: #include<bits/stdc++.h> using namespace std; #define ll long long #define pb push_back #define mem(a,b) memset(a,b,sizeof(a) ; ll dp[N]={}; ll dp1[N]={}; ll a[N]; ll t[N]={}; ll t1[N]={}; int main() { ios::sync_with_st…

题目链接:http://codeforces.com/contest/828/problem/D 题解:任意去一个点为根然后有几个k就是几个子叶也就是根结点有几个分支然后最好的解法就是贪心,将剩下的点均匀的分到各个分支上就行具体看一下代码. #include <iostream> #include <cstring> #include <cstdio> using namespace std; const int M = 2e5 + 10; int main() { i…