Codeforces Round #797 (Div. 3) E. Price Maximization

Codeforces Round #797 (Div. 3) E. Price Maximization

Note that we do not need to consider the numbers $x\ge k$, we are only interested in the remainder of the division of $x$ by $k$, and we simply add the value ⌊$\frac{x}{k}$⌋ to the answer.

We get an array $a$, where $a_i<k$. Let’s sort it and greedily type index pairs $i<j$ such that $ai+aj\ge k$. This can be done with two pointers. Then add the number of matching pairs to the answer counter. This will be the answer to the problem.

#include
using namespace std;

int main()
{
    int T;cin>>T;
    while(T--)
    {
        int n,k;cin>>n>>k;
        multiset<int> b;
        long long ans=0;
        for(int i=0;i<n;i++)
        {
            int x;cin>>x;
            ans+=x/k;
            b.insert(x%k);
        }
        while(!b.empty())
        {
            int x=*b.begin();
            b.erase(b.begin());

            auto it=b.lower_bound(k-x);
            if(it!=b.end())
            {
                b.erase(it);
                ans++;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}
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