C. Dishonest Sellers
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Igor found out discounts in a shop and decided to buy n items. Discounts at the store will last for a week and Igor knows about each item that its price now is ai, and after a week of discounts its price will be bi.

Not all of sellers are honest, so now some products could be more expensive than after a week of discounts.

Igor decided that buy at least k of items now, but wait with the rest of the week in order to save money as much as possible. Your task is to determine the minimum money that Igor can spend to buy all n items.

Input

In the first line there are two positive integer numbers n and k (1 ≤ n ≤ 2·105, 0 ≤ k ≤ n) — total number of items to buy and minimal number of items Igor wants to by right now.

The second line contains sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 104) — prices of items during discounts (i.e. right now).

The third line contains sequence of integers b1, b2, ..., bn (1 ≤ bi ≤ 104) — prices of items after discounts (i.e. after a week).

Output

Print the minimal amount of money Igor will spend to buy all n items. Remember, he should buy at least k items right now.

Examples
input
3 1
5 4 6
3 1 5
output
10
input
5 3
3 4 7 10 3
4 5 5 12 5
output
25
Note

In the first example Igor should buy item 3 paying 6. But items 1 and 2 he should buy after a week. He will pay 3 and 1 for them. So in total he will pay 6 + 3 + 1 = 10.

In the second example Igor should buy right now items 1, 2, 4 and 5, paying for them 3, 4, 10 and 3, respectively. Item 3 he should buy after a week of discounts, he will pay 5 for it. In total he will spend 3 + 4 + 10 + 3 + 5 = 25.

思路;

  计算价值然后排序水过(至少k个不是共k个);

来,上代码:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm> #define maxn 200005 using namespace std; struct SortNodeType {
int ai,bi,vi;
};
struct SortNodeType item[maxn]; int if_z,n,k,ans; char Cget; inline void in(int &now)
{
now=,if_z=,Cget=getchar();
while(Cget>''||Cget<'')
{
if(Cget=='-') if_z=-;
Cget=getchar();
}
while(Cget>=''&&Cget<='')
{
now=now*+Cget-'';
Cget=getchar();
}
now*=if_z;
} bool cmp(struct SortNodeType a,struct SortNodeType b)
{
if(a.vi!=b.vi) return a.vi<b.vi;
else
{
return a.ai<b.ai;
}
} int main()
{
in(n),in(k);
for(int i=;i<=n;i++) in(item[i].ai);
for(int i=;i<=n;i++)
{
in(item[i].bi);
item[i].vi=item[i].ai-item[i].bi;
}
sort(item+,item+n+,cmp);
int pos=;
for(int i=;i<=n;i++)
{
if(item[i].vi<=) pos++;
else break;
}
k=max(k,pos);
for(int i=;i<=k;i++) ans+=item[i].ai;
for(int i=k+;i<=n;i++) ans+=item[i].bi;
cout<<ans;
return ;
}

AC日记——Dishonest Sellers Codeforces 779c的更多相关文章

  1. AC日记——Cards Sorting codeforces 830B

    Cards Sorting 思路: 线段树: 代码: #include <cstdio> #include <cstring> #include <iostream> ...

  2. AC日记——Card Game codeforces 808f

    F - Card Game 思路: 题意: 有n张卡片,每张卡片三个值,pi,ci,li: 要求选出几张卡片使得pi之和大于等于给定值: 同时,任意两两ci之和不得为素数: 求选出的li的最小值,如果 ...

  3. AC日记——Success Rate codeforces 807c

    Success Rate 思路: 水题: 代码: #include <cstdio> #include <cstring> #include <iostream> ...

  4. AC日记——T-Shirt Hunt codeforces 807b

    T-Shirt Hunt 思路: 水题: 代码: #include <cstdio> #include <cstring> #include <iostream> ...

  5. AC日记——Magazine Ad codeforces 803d

    803D - Magazine Ad 思路: 二分答案+贪心: 代码: #include <cstdio> #include <cstring> #include <io ...

  6. AC日记——Broken BST codeforces 797d

    D - Broken BST 思路: 二叉搜索树: 它时间很优是因为每次都能把区间缩减为原来的一半: 所以,我们每次都缩减权值区间. 然后判断dis[now]是否在区间中: 代码: #include ...

  7. AC日记——Array Queries codeforces 797e

    797E - Array Queries 思路: 分段处理: 当k小于根号n时记忆化搜索: 否则暴力: 来,上代码: #include <cmath> #include <cstdi ...

  8. AC日记——Maximal GCD codeforces 803c

    803C - Maximal GCD 思路: 最大的公约数是n的因数: 然后看范围k<=10^10; 单是答案都会超时: 但是,仔细读题会发现,n必须不小于k*(k+1)/2: 所以,当k不小于 ...

  9. AC日记——Vicious Keyboard codeforces 801a

    801A - Vicious Keyboard 思路: 水题: 来,上代码: #include <cstdio> #include <cstring> #include < ...

随机推荐

  1. Golang 简单静态web服务器

    直接使用 net.http 包,非常方便 // staticWeb package main import ( "fmt" "net/http" "s ...

  2. Ubuntu 18.04 下用命令行安装Sublime

    介绍: 添加来源: $ wget -qO - https://download.sublimetext.com/sublimehq-pub.gpg | sudo apt-key add - $ sud ...

  3. 08GNU as汇编

    1. 概述 ​ 由于操作系统许多关键代码要求有很高的执行速度和效率,因此在一个操作系统源代码中通常就会包含大约 10% 左右的起关键作用的汇编语言程序量.Linux 操作系统也不例外,它的 32 位初 ...

  4. 我的Python分析成长之路7

    类 一.编程范式: 1.函数式编程   def 2.面向过程编程   (Procedural Programming) 基本设计思路就是程序一开始是要着手解决一个大的问题,然后把一个大问题分解成很多个 ...

  5. Oracle redo与undo 第一弹

      一. 什么是redo(用于前滚数据) redo也就是重做日志文件(redo log file),Oracle维护着两类重做日志文件:在线(online)重做日志文件和归档(archived)重做日 ...

  6. Linux交换分区swap

    一.SWAP 说明 1.1 SWAP 概述 当系统的物理内存不够用的时候,就需要将物理内存中的一部分空间释放出来,以供当前运行的程序使用.那些被释放的空间可能来自一些很长时间没有什么操作的程序,这些被 ...

  7. CF 510b Fox And Two Dots

    Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on ...

  8. 【HIHOCODER 1599】逃离迷宫4

    描述 小Hi被坏女巫抓进一座由无限多个格子组成的矩阵迷宫. 小Hi一开始处于迷宫(x, y)的位置,迷宫的出口在(a, b).小Hi发现迷宫被女巫施加了魔法,假设当前他处在(x, y)的位置,那么他只 ...

  9. POJ:1753-Flip Game(二进制+bfs)

    题目链接:http://poj.org/problem?id=1753 Flip Game Time Limit: 1000MS Memory Limit: 65536K Description Fl ...

  10. HTTP认证之基本认证——Basic(二)

    导航 HTTP认证之基本认证--Basic(一) HTTP认证之基本认证--Basic(二) HTTP认证之摘要认证--Digest(一) HTTP认证之摘要认证--Digest(二) 在HTTP认证 ...