Dishonest Sellers
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.
题意：给你 n件衣服 降价前的价格 以及降价后的价格 （降价，不一定价格降低），降价是在一周后，然后 你必须在 降价前至少买 k 件衣服，剩下的衣服 在 降价后买。 问最后买完这 n 件衣服所需要的最少的钱。
思路：贪心，贪心策略... 把一周后价钱比现在贵的衣服 全部买了。 如果 还未达到 k 件，就依次 再买价格降得少的，买到 k 件为止。
AC代码：
#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
struct node{
int x,y;
}item[maxn];
bool cmp(node A,node B){
return A.x - A.y < B.x - B.y;
}
int main()
{
int n,k;
while(~scanf("%d%d",&n,&k)){
for(int i = 1;i <= n;i ++)
scanf("%d",&item[i].x);
for(int i = 1;i <= n;i ++)
scanf("%d",&item[i].y);
sort(item + 1,item + 1 + n,cmp);
long long ans = 0;
for(int i = 1;i <= k;i ++)
ans += item[i].x;
int m = k + 1;
while(item[m].x - item[m].y < 0){
ans += item[m].x;
m ++;
}
for(int i = m;i <= n;i ++)
ans += item[i].y;
printf("%I64d\n",ans);
}
return 0;
}