Codeforces Round #664 (Div. 2)
D
把数按与 (m) 的大小分为两类。假设我们选择了 (i) 个不超过 (m) 的数，那么就可以选 (lceil frac{n-i-1}{d+1} ceil + 1) 个超过 (m) 的数。枚举 (i) 即可求出最大值。
Talk is cheap.Show me the code.
#include<bits/stdc++.h>
#define int long long
using namespace std;
inline int read() {
int x=0,f=1; char ch=getchar();
while(ch<'0' || ch>'9') { if(ch=='-') f=-1; ch=getchar(); }
while(ch>='0'&&ch<='9') { x=(x<<3)+(x<<1)+(ch^48); ch=getchar(); }
return x * f;
}
const int N = 1e5+7;
int n,d,m,ans;
int a[N],s1[N],s2[N];
bool cmp(int x,int y) {
return x > y;
}
signed main()
{
n = read(), d = read(), m = read();
for(int i=1;i<=n;++i) a[i] = read();
sort(a+1, a+1+n, cmp);
int q = 0;
for(int i=1;i<=n+1;++i)
if(a[i] <= m) {
q = i - 1; break;
}
for(int i=1;i<=q;++i) s1[i] = s1[i-1] + a[i];
for(int i=q+1;i<=n;++i) s2[i-q] = s2[i-q-1] + a[i];
for(int i=0;i<=n;++i) {
ans = max(ans,s1[(n-i-1)/(d+1)+1]+s2[i]);
}
printf("%lld
",ans);
return 0;
}
/*
5 2 11
8 10 15 23 5
48
*/