E. Yet Another Division Into Teams There are n students at your university. The programming skill of the i-th student is ai. As a coach, you want to divide them into teams to prepare them for the upcoming ICPC finals. Just imagine how good this unive…

Codeforces Round #598 (Div. 3)- E. Yet Another Division Into Teams - 动态规划 [Problem Description] 给你\(n\)个数,将其划分为多组,对于每个组定义其\(d\)值为 组内的最大值减最小值,问如何划分使得最终所有组的\(d\)值之和最小.每个组至少要保证有\(3\)个数. [Solution] 将所有值从小到大排序,然后我们知道最多有\(5\)个人划分到同一组中,如果有\(6\)个人,那么划分为两组一定比…

E. Yet Another Division Into Teams 首先要想明白一个东西,就是当一个小组达到六个人的时候,它一定可以拆分成两个更优的小组. 这个题可以用动态规划来写,用一个数组来保存状态,用一个队列来尝试新的状态,但是因为上面的这个特性,每一次只会有三个新的状态. 我们用sum来保存躲避选择的元素,举个例子: 分组情况为:1 2 3 | 5 6 8 11 | 20 21 22 (不一定满足题意,只是为了说明sum的意义) 则sum=(5-3)+(20-11)=11 那么这样分组…

题意 n个人,每人有一个能力值a[i],要求分成多个队伍,每个队伍至少3个人,使得所有队伍的max(a[i])-min(a[i])之和最小. 分析 不会巧妙的dp,想了一天只想到了暴力的dp. 先排序,设\(dp[i]\)表示到前i个数组队,所有队伍的最小极差和. 转移方程为\(dp[i]=min(dp[j-1]+a[i]-a[j])\)for j in 1...i-2.即\(dp[i]=a[i]+min(dp[j-1]-a[j])\). 所以可以枚举i,然后用优先队列维护\(dp[j-1]-a…

1. Massachusetts Institute of Technology, Cambridge, MA Massachusetts Institute of Technology is a private institution that was founded in 1861. It has a total undergraduate enrollment of 4,299, its setting is urban, and the campus size is 168 acres.…

B. Coach 题目连接: http://www.codeforces.com/contest/300/problem/A Description A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive. Before the university progra…

Description A programming coach has n students to teach. We know that n is divisible by 3. Let's assume that all students are numbered from 1 to n, inclusive. Before the university programming championship the coach wants to split all students into g…

这一章介绍数据绑定.本章共计27个示例,全都在VS2008下.NET3.5测试通过,点击这里下载:ConnectedData.rar 1．ShowDataWithoutBinding注: <?Mapping XmlNamespace="local" ClrNamespace="TestBinding" ?> 语法已经升级为: xmlns:local="clr-namespace:TestBinding" 这个例子讲的是在WPF中使用传…

传送门 A. Payment Without Change 签到. Code /* * Author: heyuhhh * Created Time: 2019/11/4 21:19:19 */ #include <bits/stdc++.h> #define MP make_pair #define fi first #define se second #define sz(x) (int)(x).size() #define all(x) (x).begin(), (x).end() #d…

CF598:div3解题报告 A: Payment Without Change 思路: 按题意模拟即可. 代码: #include<bits/stdc++.h> using namespace std; typedef long long ll; int main() { ll T; cin >> T; while(T--) { ll a, b, n, s; cin >> a >> b >> n >> s; if(a*n + b &…