Codeforces Round #267 (Div. 2) C. George and Job(DP)补题
Codeforces Round #267 (Div. 2) C. George and Job题目链接请点击~
The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the following problem at the work.
Given a sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers:
[l1, r1], [l2, r2], ..., [lk, rk] (1 ≤ l1 ≤ r1 < l2 ≤ r2 < ... < lk ≤ rk ≤ n; ri - li + 1 = m),
in such a way that the value of sum is maximal possible. Help George to cope with the task.
The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integersp1, p2, ..., pn (0 ≤ pi ≤ 109).
Print an integer in a single line — the maximum possible value of sum.
5 2 1
1 2 3 4 5
9
7 1 3
2 10 7 18 5 33 0
61
#include <iostream>
#include <cstdio>
#define LL long long
using namespace std;
const int maxn = + ;
LL p[maxn],s[maxn],dp[maxn][maxn];
int main(){
int n,m,k;
cin>>n>>m>>k;
for(int i = ;i <= n;i++){
cin>>p[i];s[i] = s[i - ] + p[i];
}
for(int i = m;i <= n;i++)
for(int j = ;j <= k;j++)
dp[i][j] = max(dp[i-m][j-]+s[i]-s[i-m],dp[i-][j]);
cout<<dp[n][k]<<endl;
return ;
}
Codeforces Round #267 (Div. 2) C. George and Job(DP)补题的更多相关文章
- Codeforces Round #267 (Div. 2) C. George and Job DP
C. George and Job The new ITone 6 has been released ...
- 01背包 Codeforces Round #267 (Div. 2) C. George and Job
题目传送门 /* 题意:选择k个m长的区间,使得总和最大 01背包:dp[i][j] 表示在i的位置选或不选[i-m+1, i]这个区间,当它是第j个区间. 01背包思想,状态转移方程:dp[i][j ...
- Codeforces Round #267 (Div. 2) C. George and Job (dp)
wa哭了,,t哭了,,还是看了题解... 8170436 2014-10-11 06:41:51 njczy2010 C - George and Jo ...
- Codeforces Round #367 (Div. 2) C. Hard problem(DP)
Hard problem 题目链接: http://codeforces.com/contest/706/problem/C Description Vasiliy is fond of solvin ...
- Codeforces Round #368 (Div. 2) A. Brain's Photos (水题)
Brain's Photos 题目链接: http://codeforces.com/contest/707/problem/A Description Small, but very brave, ...
- Codeforces Round #227 (Div. 2) E. George and Cards set内二分+树状数组
E. George and Cards George is a cat, so he loves playing very much. Vitaly put n cards in a row in ...
- Codeforces Round #227 (Div. 2) E. George and Cards 线段树+set
题目链接: 题目 E. George and Cards time limit per test:2 seconds memory limit per test:256 megabytes 问题描述 ...
- Codeforces Round #267 (Div. 2) A
题目: A. George and Accommodation time limit per test 1 second memory limit per test 256 megabytes inp ...
- Codeforces Round #267 (Div. 2) B. Fedor and New Game【位运算/给你m+1个数让你判断所给数的二进制形式与第m+1个数不相同的位数是不是小于等于k,是的话就累计起来】
After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play ...
随机推荐
- 笔试算法题(34):从数字序列中寻找仅出现一次的数字 & 最大公约数(GCD)问题
出题:给定一个数字序列,其中每个数字最多出现两次,只有一个数字仅出现了一次,如何快速找出其中仅出现了一次的数字: 分析: 由于知道一个数字异或操作它本身(X^X=0)都为0,而任何数字异或操作0都为它 ...
- [LUOGU] P1466 集合 Subset Sums
题目描述 对于从1到N (1 <= N <= 39) 的连续整数集合,能划分成两个子集合,且保证每个集合的数字和是相等的.举个例子,如果N=3,对于{1,2,3}能划分成两个子集合,每个子 ...
- Java权限管理(授权与认证)
CRM权限管理 有兴趣的同学也可以阅读我最近分享的:Shiro框架原理分析 (PS : 这篇博客里面介绍了使用Shiro框架的方式实现权限管理) https://www.cnblogs.com/y ...
- nodejs学习(二) ---- express中使用模板引擎jade
系列教程,上一节教程 express+nodejs快速创建一个项目 在创建一个项目后,views目录下的文件后缀为 .jade . 打开 index.jade,具体内容如下图(忽略 header.j ...
- assert.notDeepEqual()
assert.notDeepEqual(actual, expected[, message]) 深度地不相等比较测试,与 assert.deepEqual() 相反. const assert = ...
- 13Spring通过注解配置Bean(1)
配置Bean的形式:基于XML文件的方式:基于注解的方式(基于注解配置Bean:基于注解来装配Bean的属性) 下面介绍基于注解的方式来配置Bean. ——组件扫描(component scannin ...
- C++动态申请内存 new T()与new T[]的区别
new与delete 我们知道,new和delete运算符是用于动态分配和撤销内存的运算符. new的用法 开辟单变量地址空间: i. 如 new int ; 指开辟一个存放数组的存储空间,返回一个指 ...
- 如何用nfs命令烧写内核和文件系统(网络下载文件到nandflash)(未完)
使用tftp下载烧写 a.设uboot里的ip地址 set ipaddr 192.168.1.17(uboot的ip设置成同网段) set serverip 192.168.1.5(电脑本机作为服务i ...
- JavaScript编程那些事(牛客网 LeetCode)
计算给定数组 arr 中所有元素的总和 本人提供常规方法 function sum(arr) { var len = arr.length; var sum = 0; if(len == 0){ su ...
- Quartz.Net 学习之路02 初探Quartz.Net
第二讲:简单的实例,看看Quartz.Net强在哪里? 直接上代码,代码里有注释: using System; using Quartz; using Quartz.Impl; namespace L ...