Codeforces Round #376 (Div. 2) F. Video Cards 数学,前缀和
1 second
256 megabytes
standard input
standard output
Little Vlad is fond of popular computer game Bota-2. Recently, the developers announced the new add-on named Bota-3. Of course, Vlad immediately bought only to find out his computer is too old for the new game and needs to be updated.
There are n video cards in the shop, the power of the i-th video card is equal to integer value ai. As Vlad wants to be sure the new game will work he wants to buy not one, but several video cards and unite their powers using the cutting-edge technology. To use this technology one of the cards is chosen as the leading one and other video cards are attached to it as secondary. For this new technology to work it's required that the power of each of the secondary video cards is divisible by the power of the leading video card. In order to achieve that the power of any secondary video card can be reduced to any integer value less or equal than the current power. However, the power of the leading video card should remain unchanged, i.e. it can't be reduced.
Vlad has an infinite amount of money so he can buy any set of video cards. Help him determine which video cards he should buy such that after picking the leading video card and may be reducing some powers of others to make them work together he will get the maximum total value of video power.
The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of video cards in the shop.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 200 000) — powers of video cards.
The only line of the output should contain one integer value — the maximum possible total power of video cards working together.
4
3 2 15 9
27
4
8 2 2 7
18
In the first sample, it would be optimal to buy video cards with powers 3, 15 and 9. The video card with power 3 should be chosen as the leading one and all other video cards will be compatible with it. Thus, the total power would be 3 + 15 + 9 = 27. If he buys all the video cards and pick the one with the power 2 as the leading, the powers of all other video cards should be reduced by 1, thus the total power would be 2 + 2 + 14 + 8 = 26, that is less than 27. Please note, that it's not allowed to reduce the power of the leading video card, i.e. one can't get the total power 3 + 1 + 15 + 9 = 28.
In the second sample, the optimal answer is to buy all video cards and pick the one with the power 2 as the leading. The video card with the power 7 needs it power to be reduced down to 6. The total power would be 8 + 2 + 2 + 6 = 18.
题意:给你n个数,max((i=1-n)sigma(j=1-n)(a[j]-a[j]%a[i]);
思路:枚举每个数,再枚举它的倍数,利用前缀和得到(base *k- base*(k+1))之间的个数;复杂度o(nlogn);
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=2e5+,M=1e6+,inf=1e9+,mod=1e9+;
const ll INF=1e18+;
int sum[N],a[N];
int main()
{
int n,x;
scanf("%d",&n);
for(int i=;i<=n;i++)
scanf("%d",&x),a[x]++;
for(int i=;i<=;i++)sum[i]=sum[i-]+a[i];
ll ans=;
for(int i=;i<=;i++)
{
if(a[i])
{
ll s=;
for(int j=;j<=;j+=i)
{
int nex=min(,j+i-);
if(j!=)
s+=(ll)(sum[nex]-sum[j-])*j;
}
ans=max(ans,s);
}
}
printf("%lld\n",ans);
return ;
}
Codeforces Round #376 (Div. 2) F. Video Cards 数学,前缀和的更多相关文章
- Codeforces Round #376 (Div. 2)F. Video Cards(前缀和)
题目链接:http://codeforces.com/contest/731/problem/F 题意:有n个数,从里面选出来一个作为第一个,然后剩下的数要满足是这个数的倍数,如果不是,只能减小为他的 ...
- Codeforces Round #376 (Div. 2) F. Video Cards 数学 & 暴力
http://codeforces.com/contest/731/problem/F 注意到一个事实,如果你要找一段区间中(从小到大的),有多少个数是能整除左端点L的,就是[L, R]这样.那么,很 ...
- Codeforces Round #376 (Div. 2) F. Video Cards —— 前缀和 & 后缀和
题目链接:http://codeforces.com/contest/731/problem/F F. Video Cards time limit per test 1 second memory ...
- Codeforces Round #585 (Div. 2) A. Yellow Cards(数学)
链接: https://codeforces.com/contest/1215/problem/A 题意: The final match of the Berland Football Cup ha ...
- Codeforces Round #485 (Div. 2) F. AND Graph
Codeforces Round #485 (Div. 2) F. AND Graph 题目连接: http://codeforces.com/contest/987/problem/F Descri ...
- Codeforces Round #486 (Div. 3) F. Rain and Umbrellas
Codeforces Round #486 (Div. 3) F. Rain and Umbrellas 题目连接: http://codeforces.com/group/T0ITBvoeEx/co ...
- Codeforces Round #501 (Div. 3) F. Bracket Substring
题目链接 Codeforces Round #501 (Div. 3) F. Bracket Substring 题解 官方题解 http://codeforces.com/blog/entry/60 ...
- Codeforces Round #499 (Div. 1) F. Tree
Codeforces Round #499 (Div. 1) F. Tree 题目链接 \(\rm CodeForces\):https://codeforces.com/contest/1010/p ...
- Codeforces 731 F. Video Cards(前缀和)
Codeforces 731 F. Video Cards 题目大意:给一组数,从中选一个数作lead,要求其他所有数减少为其倍数,再求和.问所求和的最大值. 思路:统计每个数字出现的个数,再做前缀和 ...
随机推荐
- divcss5布局
一.ie9不支持line-height字体垂直居中兼容问题 原因:CSS中使用了中文字体,而中文字体使用汉字.如:font-family:"微软雅黑" 1.将中文字体汉字 ...
- java中length,length(),size()的区别
1. java中的length属性是针对数组说的,比如说你声明了一个数组,想知道这个数组的长度则用到了length这个属性.2. java中的length()方法是针对字符串String说的,如果想看 ...
- Index Condition Pushdown Optimization
Index Condition Pushdown (ICP) is an optimization for the case where MySQL retrieves rows from a tab ...
- ch1:python3 查看版本号、安装目录和工作空间目录
查看python版本: 每次打开python顶端会显示版本号 在程序中判断版本号可以通过import sys sys.version 在dos下可以通过python -V查看 安装目录:C:\Pyt ...
- Web API 和 WCF 的比较
现在有很多可用的技术允许你创建被不同客户端所消费的服务,这些客户端可能是Web应用程序.Windows应用程序和移动应用等.服务可以支持http协议或者其他协议.接下来的讨论仅限于ASP.NET We ...
- 4.1HTML和Bootstrap css精华
1.HTML 2.理解Bootstrap HTML元素告诉浏览器,他要表现的是什么类型的内容,当他们不提供任何关于如何显示内容的信息.如何显示内容的信息,由CSS提供. 本书仅包含足够的信息,让你查看 ...
- phpize 动态添加 PHP 扩展的错误及解决方案
使用phpize 动态添加 PHP 扩展是开发中经常需要做的事情,但是在 macOS 中,首次使用该功能必然会碰到一些错误,本文列出了这些错误的解决方法. 问题一: 执行 phpize 报错如下: $ ...
- 15、Jdbc的优化(BeanUtils组件)
Jdbc的优化! BeanUtils组件 自定义一个持久层的框架 DbUtils组件 案例优化 1. BeanUtils组件 1.1 简介 程序中对javabean的操作很频繁, 所以apach ...
- 如何修改ECShop发货单查询显示个数
使用ecshop的朋友都知道,商城首页调用的发货单查询,默认显示的10个.很多朋友想修改它的数量,可是在后台管理却找不到相应的地方,这个修改和显示排行榜的数量修改方法不一样.排行榜是可以在后台修改的, ...
- 一个很不错的bash脚本编写教程
转自 http://blog.chinaunix.net/uid-20328094-id-95121.html 一个很不错的bash脚本编写教程,至少没接触过BASH的也能看懂! 建立一个脚本 Lin ...