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. webpack4搭建Vue开发环境笔记~~持续更新

    项目git地址 一.node知识 __dirname: 获取当前文件所在路径,等同于path.dirname(__filename) console.log(__dirname); // Prints ...

  2. python-函数的对象、函数嵌套、名称空间和作用域

    目录 函数的对象 函数对象的四大功能 引用 当做参数传给一个函数 可以当做函数的返回值 可以当做容器类型的元素 函数的嵌套 函数的嵌套定义 函数的嵌套调用 名称空间与作用域 名称空间 内置名称空间 全 ...

  3. python 列表加法"+"和"extend"的区别

    相同点 : "+"和"extend"都能将两个列表成员拼接到到一起 不同点 :   + : 生成的是一个新列表(id改变) extend : 是将一个列表的成员 ...

  4. perl-basic-数据类型&引用

    我觉得这一系列的标题应该是:PERL,从入门到放弃 USE IT OR U WILL LOSE IT 参考资料: https://qntm.org/files/perl/perl.html 在线per ...

  5. FSMC原理通俗解释

    所以不用GPIO口直接驱动液晶,是因为这种方法速度太慢,而FSMC是用来外接各种存储芯片的,所以其数据通信速度是比普通GPIO口要快得多的.TFT-LCD 驱动芯片的读写时序和SRAM的差不多,所以就 ...

  6. TypeError: cannot use a string pattern on a bytes-like object

    一劳永逸解决:TypeError: cannot use a string pattern on a bytes-like object TypeError: cannot use a string ...

  7. SQL_4_函数

    在SQL的函数中可以执行一些诸如对某一些进行汇总或将一个字符串中的字符转换为大写的操作等: 函数有:汇总函数.日期与时间函数.数学函数.字符函数.转换函数与其他函数. 汇总函数 这是一组函数,它们返回 ...

  8. 启动Chrome浏览器弹出“You are using an unsupported command-line flag –ignore-certificate-errors. Stability and security will suffer”

    采用如下代码: public static void launchChrome() { System.setProperty("webdriver.chrome.driver", ...

  9. BZOJ 4057: [Cerc2012]Kingdoms

    状压DP #include<cstdio> #include<cstring> using namespace std; int F[1200005],A[25][25],st ...

  10. UVa 10118 记忆化搜索 Free Candies

    假设在当前状态我们第i堆糖果分别取了cnt[i]个,那么篮子里以及口袋里糖果的个数都是可以确定下来的. 所以就可以使用记忆化搜索. #include <cstdio> #include & ...