E. 我要打 k 个

合集 - 贪心(11)

1.P2887 [USACO07NOV] Sunscreen G2024-03-282.F. 0, 1, 2, Tree!2024-03-293.P2949 [USACO09OPEN] Work Scheduling G2024-05-204.C. Given Length and Sum of Digits...2024-06-065.C. Lexicographically Largest2024-06-176.B. Neutral Tonality2024-06-177.F. Lisa and the Martians2024-06-17

8.E. 我要打 k 个2024-06-18

9.D. Armchairs2024-06-1810.D. Andrey and Escape from Capygrad2024-07-1311.Apartments2024-07-30

收起

原题链接

题解

1.操作1肯定选最小的
2.操作2不在乎数的大小
3.因此我们可以优先实施操作1，然后再实施操作2

code

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+5;
ll a[N];
ll n,m,k,x;

int main(){
	cin>>n>>m>>k>>x;
	for (int i=1;i<=n;i++)  cin>>a[i];

	sort(a+1,a+n+1);
	ll ans=min((m/x),n/k)*k;
	for(ll i=1;i<=n;i++)
    {
        m-=a[i];
        if(m<0) break;
        ans=max(ans,i+min((m/x),(n-i)/k)*k);
    }

    cout<<ans;
	return 0;
}