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  • Codeforces Round #664 (Div. 2)练习

    Codeforces Round #664 (Div. 2)

    D

    题目地址

    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
    */
    
    
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  • 原文地址:https://www.cnblogs.com/BaseAI/p/13522980.html
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