B. Mischievous Mess Makers

题目连接:

http://www.codeforces.com/contest/655/problem/B

Description

It is a balmy spring afternoon, and Farmer John's n cows are ruminating about link-cut cacti in their stalls. The cows, labeled 1 through n, are arranged so that the i-th cow occupies the i-th stall from the left. However, Elsie, after realizing that she will forever live in the shadows beyond Bessie's limelight, has formed the Mischievous Mess Makers and is plotting to disrupt this beautiful pastoral rhythm. While Farmer John takes his k minute long nap, Elsie and the Mess Makers plan to repeatedly choose two distinct stalls and swap the cows occupying those stalls, making no more than one swap each minute.

Being the meticulous pranksters that they are, the Mischievous Mess Makers would like to know the maximum messiness attainable in the k minutes that they have. We denote as pi the label of the cow in the i-th stall. The messiness of an arrangement of cows is defined as the number of pairs (i, j) such that i < j and pi > pj.

Input

The first line of the input contains two integers n and k (1 ≤ n, k ≤ 100 000) — the number of cows and the length of Farmer John's nap, respectively.

Output

Output a single integer, the maximum messiness that the Mischievous Mess Makers can achieve by performing no more than k swaps.

Sample Input

5 2

Sample Output

10

Hint

题意

给你1到n的序列,然后你可以最多交换k次

让你使得逆序数最多,问你答案是多少

题解:

贪心,第一个数和最后一个数交换,第二个数和倒数第二个数交换,然后这样就好了

每次对答案的贡献是2*(n-i-i)+1

这个画画图就知道了。

代码

#include<bits/stdc++.h>
using namespace std; long long ans,n,k;
int main()
{
cin>>n>>k;
for(int i=1;i+i<=n&&i<=k;i++)
ans+=2*(n-i-i)+1;
cout<<ans<<endl;
}

CROC 2016 - Elimination Round (Rated Unofficial Edition) B. Mischievous Mess Makers 贪心的更多相关文章

  1. CROC 2016 - Elimination Round (Rated Unofficial Edition) D. Robot Rapping Results Report 二分+拓扑排序

    D. Robot Rapping Results Report 题目连接: http://www.codeforces.com/contest/655/problem/D Description Wh ...

  2. CROC 2016 - Elimination Round (Rated Unofficial Edition) D. Robot Rapping Results Report 拓扑排序+二分

    题目链接: http://www.codeforces.com/contest/655/problem/D 题意: 题目是要求前k个场次就能确定唯一的拓扑序,求满足条件的最小k. 题解: 二分k的取值 ...

  3. CROC 2016 - Elimination Round (Rated Unofficial Edition) F - Cowslip Collections 数论 + 容斥

    F - Cowslip Collections http://codeforces.com/blog/entry/43868 这个题解讲的很好... #include<bits/stdc++.h ...

  4. CROC 2016 - Elimination Round (Rated Unofficial Edition) E - Intellectual Inquiry dp

    E - Intellectual Inquiry 思路:我自己YY了一个算本质不同子序列的方法, 发现和网上都不一样. 我们从每个点出发向其后面第一个a, b, c, d ...连一条边,那么总的不同 ...

  5. CROC 2016 - Elimination Round (Rated Unofficial Edition) E. Intellectual Inquiry 贪心 构造 dp

    E. Intellectual Inquiry 题目连接: http://www.codeforces.com/contest/655/problem/E Description After gett ...

  6. CROC 2016 - Elimination Round (Rated Unofficial Edition) C. Enduring Exodus 二分

    C. Enduring Exodus 题目连接: http://www.codeforces.com/contest/655/problem/C Description In an attempt t ...

  7. CROC 2016 - Elimination Round (Rated Unofficial Edition) A. Amity Assessment 水题

    A. Amity Assessment 题目连接: http://www.codeforces.com/contest/655/problem/A Description Bessie the cow ...

  8. CF #CROC 2016 - Elimination Round D. Robot Rapping Results Report 二分+拓扑排序

    题目链接:http://codeforces.com/contest/655/problem/D 大意是给若干对偏序,问最少需要前多少对关系,可以确定所有的大小关系. 解法是二分答案,利用拓扑排序看是 ...

  9. 8VC Venture Cup 2016 - Elimination Round D. Jerry's Protest 暴力

    D. Jerry's Protest 题目连接: http://www.codeforces.com/contest/626/problem/D Description Andrew and Jerr ...

随机推荐

  1. Java多线程学习(七)并发编程中一些问题

    本节思维导图: 关注微信公众号:"Java面试通关手册" 回复"Java多线程"获取思维导图源文件和思维导图软件. 多线程就一定好吗?快吗?? 并发编程的目的就 ...

  2. python算法之近似熵、互近似熵算法

    理论基础 近似熵? 定义:近似熵是一个随机复杂度,反应序列相邻的m个点所连成折线段的模式的互相近似的概率与由m+1个点所连成的折线段的模式相互近似的概率之差. 作用:用来描述复杂系统的不规则性,越是不 ...

  3. makefile初步制作,arm-linux- (gcc/ld/objcopy/objdump)详解【转】

    转自:http://www.cnblogs.com/lifexy/p/7065175.html 在linux中输入vi Makefile 来实现创建Makefile文件 注意:命令行前必须加TAB键 ...

  4. MGR_ERROR 3092 (HY000): DROP DATABASE failed;

    start group_replication;时报以下错: ERROR 3092 (HY000): DROP DATABASE failed; some tables may have been d ...

  5. Ural Sport Programming Championship 2015

    Ural Sport Programming Championship 2015 A - The First Day at School 题目描述:给出课程安排,打印一个课程表. solution 暴 ...

  6. oracle命令生成AWR报告

    --命令生成AWR报告oracle@linux:~> sqlplus / as sysdba SQL*Plus: Release 11.1.0.7.0 - Production on Fri A ...

  7. css3动画详解

    一.Keyframes介绍: Keyframes被称为关键帧,其类似于Flash中的关键帧.在CSS3中其主要以“@keyframes”开头,后面紧跟着是动画名称加上一对花括号“{…}”,括号中就是一 ...

  8. 关于boost 的smart_ptr 的使用问题

    boost 的smart_ptr 库中含有好几种智能指针,大家用的最多的应该是shared_ptr ,为啥呢?好用,不用管他啥时候会自动删除等等,而且拷贝和复制都很到位, 但实际上,这个库也有问题,连 ...

  9. 《深入理解Java虚拟机》笔记--第十三章、线程安全与锁优化

    先保证并发的正确性,然后在此基础上来实现高效. 线程安全:     当多个线程访问一个对象时,如果不考虑这些线程在运行时环境下的调度和交替执行,也不需要进行额外的同步,或者在调用方进行任何其他的协调操 ...

  10. C/C++——C语言数组名与指针

    版权声明:原创文章,转载请注明出处. 1. 一维数组名与指针 对于一维数组来说,数组名就是指向该数组首地址的指针,对于: ]; array就是该数组的首地址,如果我们想定义一个指向该数组的指针,我们可 ...