Description

Once upon a time in a country far away lived a king and he had a big kingdom. He was a very clever king but he had one weakness -- he could count only up to three.         Nevertheless, he did not consider this to be a really great drawback, since he had a lot of wizards who could count up to one hundred (and some of them, people said, even up to one thousand), so it was all right. But one day the grief came to the kingdom as the outnumbering barbarians started to approach from all sides. And the king then had to make the most important decision in his life. He had to choose which of his sons to make generals that he would send to the borders of the country to lead the army.         However, the king knew that though some of his sons were clever, just like he was, some of them were quite stupid and could only lower army spirits with their wrong decisions. More precisely, he knew about each of his sons his mental potential -- an integer number ranging from minus three to three (remember, that the king could count only up to three). He also knew that the chance of his army defeating barbarians was proportional to the sum of some powers of mental potentials of those of his sons that he would make generals (the power exponent was a positive integer number, the same for all his sons and not exceeding three either). Thus he had to choose such a combination of his sons to lead the army, that this sum would be maximal possible.         However, the king himself could not make all apropriate calculations since, for example, the second power of the number not exceeding three (which is its square) could be greater than three, and therefore he asked you, his most intellegent wizard, to solve this problem.      

Input

The first line of the input file contains the number of the sons of the king (integer number less or equal to one hundred). The second line contains the positive integer number not exceeding three, the exponent in the formula used to calculate the chance of defeating barbarians. The third line contains the list of mental potentials of king’s sons -- all integer numbers, not greater than three by their absolute value.      

Output

Output the only number -- the maximal possible chance of defeating barbarians measured as the sum described.      

Sample Input

3
3
2 -1 1

Sample Output

9

Hint

In the example above the king should choose his first and third sons to be the generals. In this case the chance to defeat barbarians, which is the sum of cubes of mental potentials of these sons, is eight plus one, that is nine.
没看明白题的意思!!!该题是求n个数的k次平方和的最大值!!注意k为偶数时,正负数的k次方都为正数,要计算在其中!!
 
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int n,b,a,s=;
cin>>n;
cin>>a;
for(int i=;i<n;i++){
cin>>b;
if(b>||(b<&&a%==))s+=pow((double)b,a);
}
cout<<s;
//system("pause");
return ;
}

E - The King的更多相关文章

  1. BZOJ 1087: [SCOI2005]互不侵犯King [状压DP]

    1087: [SCOI2005]互不侵犯King Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 3336  Solved: 1936[Submit][ ...

  2. [bzoj1087][scoi2005]互不侵犯king

    题目大意 在N×N的棋盘里面放K个国王,使他们互不攻击,共有多少种摆放方案.国王能攻击到它上下左右,以及左上 左下右上右下八个方向上附近的各一个格子,共8个格子. 思路 首先,搜索可以放弃,因为这是一 ...

  3. King's Quest —— POJ1904(ZOJ2470)Tarjan缩点

    King's Quest Time Limit: 15000MS Memory Limit: 65536K Case Time Limit: 2000MS Description Once upon ...

  4. 【状压DP】bzoj1087 互不侵犯king

    一.题目 Description 在N×N的棋盘里面放K个国王,使他们互不攻击,共有多少种摆放方案.国王能攻击到它上.下.左.右,以及左上.左下.右上.右下八个方向上附近的各一个格子,共8个格子. I ...

  5. ZOJ 2334 Monkey King

    并查集+左偏树.....合并的时候用左偏树,合并结束后吧父结点全部定成树的根节点,保证任意两个猴子都可以通过Find找到最厉害的猴子                       Monkey King ...

  6. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 K. King’s Rout

    K. King's Rout time limit per test 4 seconds memory limit per test 512 megabytes input standard inpu ...

  7. BZOJ-1087 互不侵犯King 状压DP+DFS预处理

    1087: [SCOI2005]互不侵犯King Time Limit: 10 Sec Memory Limit: 162 MB Submit: 2337 Solved: 1366 [Submit][ ...

  8. POJ1364 King

    Description Once, in one kingdom, there was a queen and that queen was expecting a baby. The queen p ...

  9. [Educational Codeforces Round 16]A. King Moves

    [Educational Codeforces Round 16]A. King Moves 试题描述 The only king stands on the standard chess board ...

  10. codeforces A. Rook, Bishop and King 解题报告

    题目链接:http://codeforces.com/problemset/problem/370/A 题目意思:根据rook(每次可以移动垂直或水平的任意步数(>=1)),bishop(每次可 ...

随机推荐

  1. respondsToSelector的相关使用

    -(BOOL) isKindOfClass: classObj 用来判断是否是某个类或其子类的实例 -(BOOL) isMemberOfClass: classObj 用来判断是否是某个类的实例 -( ...

  2. 一次$.getJSON不执行的记录

    别人的代码,拿过来调,发现修改功能都不能用,修改时通过ajax发json获取数据的,看chrome开发者工具发现有发送数据,也有返回值: 发起请求并获取数据,发现回调函数不执行! $.getJSON( ...

  3. read/load

    ready先执行,load后执行. DOM文档加载的步骤: (1) 解析HTML结构. (2) 加载外部脚本和样式表文件. (3) 解析并执行脚本代码. (4) 构造HTML DOM模型.//read ...

  4. pubwin2009服务端 修改系统时间方法

      所有操作都是必须的,包括重启. 1.建议操作前全体结账,并交班---不结帐的话, 改完时间会出问题的.2.停止pubwin2009服务端 3.将C:\WINDOWS\system32\driver ...

  5. .net 拉姆达 groupby(p => p.X) order by count(c.Count())

    //国家 var entityCountriesList = aliexpressEntities.SYS_CourierCode.Where(whereSelect.Compile()).Group ...

  6. javascript之Boolean

    一.new Boolean(value); //构造函数 Boolean(value); //转换函数 0,NaN,null,"",undefined都会转成false,其他都会返 ...

  7. Qt编程之QtScript

    需求是这样的: 需要给一个硬件接口测试些东西,用js来调用,js做成不同的独立模块分别测试不同的硬件接口,包括DMD内存,PCIE带宽等等.需要用一个exe来载入这些js文件分别调用测试模块.exe用 ...

  8. UESTC_树上战争 CDOJ 32

    给一棵树,如果树上的某个节点被某个人占据,则它的所有儿子都被占据,lxh和pfz初始时分别站在两个节点上,谁当前所在的点被另一个人占据,他就输了比赛,问谁能获胜. Input 输入包含多组数据 每组第 ...

  9. poj 1020 Anniversary Cake(切正方形蛋糕+搜索)

                                                                                                         ...

  10. HDU4762(JAVA大数)

    Cut the Cake Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tota ...