[ABC246C] Coupon

Problem Statement

There are $N$ items in a shop. For each $i = 1, 2, \ldots, N$, the price of the $i$-th item is $A_i$ yen (the currency of Japan).

Takahashi has $K$ coupons.

Each coupon can be used on one item. You can use any number of coupons, possibly zero, on the same item. Using $k$ coupons on an item with a price of $a$ yen allows you to buy it for $\max\lbrace a - kX, 0\rbrace$ yen.

Print the minimum amount of money Takahashi needs to buy all the items.

Constraints

• $1 \leq N \leq 2 \times 10^5$
• $1 \leq K, X \leq 10^9$
• $1 \leq A_i \leq 10^9$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$N$ $K$ $X$
$A_1$ $A_2$ $\ldots$ $A_N$

Output

Print the answer.

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Sample Input 1

5 4 7
8 3 10 5 13

Sample Output 1

12

By using $1$ coupon on the $1$-st item, $1$ coupon on the $3$-rd item, and $2$ coupons on the $5$-th item, Takahashi can:

• buy the $1$-st item for $\max\lbrace A_1-X, 0 \rbrace = 1$ yen,
• buy the $2$-nd item for $\max\lbrace A_2, 0 \rbrace = 3$ yen,
• buy the $3$-rd item for $\max\lbrace A_3-X, 0 \rbrace = 3$ yen,
• buy the $4$-th item for $\max\lbrace A_4, 0 \rbrace = 5$ yen,
• buy the $5$-th item for $\max\lbrace A_5-2X, 0 \rbrace = 0$ yen,

for a total of $1 + 3 + 3 + 5 + 0 = 12$ yen, which is the minimum possible.

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Sample Input 2

5 100 7
8 3 10 5 13

Sample Output 2

0

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Sample Input 3

20 815 60
2066 3193 2325 4030 3725 1669 1969 763 1653 159 5311 5341 4671 2374 4513 285 810 742 2981 202

Sample Output 3

112

发现如果减少后没有小于 0，那么可以直接减 $X$。所以我们先把能减的减掉。
剩下的，把他从大到小排序完后，尽量减前面的就可以了。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=2e5+5;
int n,k,x,a[N];
LL s,ret;
int cmp(int x,int y)
{
	return x>y;
}
int main()
{
	scanf("%d%d%d",&n,&k,&x);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",a+i);
		ret+=a[i]/x;
		a[i]%=x;
		s+=a[i];
	}
	sort(a+1,a+n+1,cmp);
	if(k<ret)
		printf("%lld",1LL*(ret-k)*x+s);
	else
	{
		for(int i=1;i<=n&&i<=k-ret;i++)
			s-=a[i],a[i]=0;
		printf("%lld",s);
	}
}