Codeforces Round #476 (Div. 2) [Thanks, Telegram!] C. Greedy Arkady(brute force/math)

这题的关键就是算出自己一共receive了多少次,我们可以很容易的便推出这么一个式子

\[
\color{black}{\frac{(\frac{n}{x}+k-1)}{k}}
\]

之后暴力二分就好了

#include<cstdio>
#include<iomanip>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<vector>
#include<set>
#include<queue>
#include<limits.h>
#include<string.h>
#include<map>
#include<list>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

#define inf int(0x3f3f3f3f)
#define mod int(1e9+7)
#define eps double(1e-6)
#define pi acos(-1.0)
#define lson  root << 1
#define rson  root << 1 | 1

ll n,k,m,d;

ll i;

ll cal(ll x)
{
    //cout<<111<<endl;
    return (n/x+k-1)/k;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin>>n>>k>>m>>d;
    ll ans=0;
    for(i=1; i<=d; i++)
    {
        ll Max=0;
        ll l=1,r=m;
        while(l<=r)
        {
            //cout<<111<<endl;
            ll mid=l+(r-l)/2;
            ll num=cal(mid);
            //cout<<111<<endl;
            if(num>i)
                l=mid+1;
            else if(num<i)
                r=mid-1;
            else
            {
                Max=i*mid;
                l=mid+1;
            }
        }
        ans=max(ans,Max);
    }
    cout<<ans<<endl;
    return 0;
}