C. Greedy Arkady

Posted 2020-11-02 planche

$\mathrm{kk people\; want\; to\; split nn candies\; between\; them.\; Each\; candy\; should\; be\; given\; to\; exactly\; one\; of\; them\; or\; be\; thrown\; away.}$

The people are numbered from $\mathrm{11 to kk,\; and\; Arkady\; is\; the\; first\; of\; them.\; To\; split\; the\; candies,\; Arkady\; will\; choose\; an\; integer xx and\; then\; give\; the\; first xx candies\; to\; himself,\; the\; next xx candies\; to\; the\; second\; person,\; the\; next xx candies\; to\; the\; third\; person\; and\; so\; on\; in\; a\; cycle.\; The\; leftover\; (the\; remainder\; that\; is\; not\; divisible\; by xx)\; will\; be\; thrown\; away.}$

Arkady can‘t choose $\mathrm{xx greater\; than MM as\; it\; is\; considered\; greedy.\; Also,\; he\; can‘t\; choose\; such\; a\; small xxthat\; some\; person\; will\; receive\; candies\; more\; than DD times,\; as\; it\; is\; considered\; a\; slow\; splitting.}$

Please find what is the maximum number of candies Arkady can receive by choosing some valid $\mathrm{xx.}$

Input

The only line contains four integers $\mathrm{nn, kk, MM and DD (2\le n\le 10182\le n\le 1018, 2\le k\le n2\le k\le n, 1\le M\le n1\le M\le n, 1\le D\le min(n,1000)1\le D\le min(n,1000), M\cdot D\cdot k\ge nM\cdot D\cdot k\ge n) —\; the\; number\; of\; candies,\; the\; number\; of\; people,\; the\; maximum\; number\; of\; candies\; given\; to\; a\; person\; at\; once,\; the\; maximum\; number\; of\; times\; a\; person\; can\; receive\; candies.}$

Output

Print a single integer — the maximum possible number of candies Arkady can give to himself.

Note that it is always possible to choose some valid $\mathrm{xx.}$

Examples

input

Copy

20 4 5 2

output

Copy

8

input

Copy

30 9 4 1

output

Copy

4

Note

In the first example Arkady should choose $\mathrm{x=4x=4.\; He\; will\; give 44 candies\; to\; himself, 44 candies\; to\; the\; second\; person, 44 candies\; to\; the\; third\; person,\; then 44 candies\; to\; the\; fourth\; person\; and\; then\; again 44candies\; to\; himself.\; No\; person\; is\; given\; candies\; more\; than 22 times,\; and\; Arkady\; receives 88 candies\; in\; total.}$

Note that if Arkady chooses $\mathrm{x=5x=5,\; he\; will\; receive\; only 55 candies,\; and\; if\; he\; chooses x=3x=3,\; he\; will\; receive\; only 3+3=63+3=6 candies\; as\; well\; as\; the\; second\; person,\; the\; third\; and\; the\; fourth\; persons\; will\; receive 33candies,\; and 22 candies\; will\; be\; thrown\; away.\; He\; can‘t\; choose x=1x=1 nor x=2x=2 because\; in\; these\; cases\; he\; will\; receive\; candies\; more\; than 22 times.}$

In the second example Arkady has to choose $\mathrm{x=4x=4,\; because\; any\; smaller\; value\; leads\; to\; him\; receiving\; candies\; more\; than 11 time.}$

 

 

数学题，d比较小，枚举d

贪心思考，最好应该是第一个人取i次，其他人取i-1次 

x*i+x*(i-1)*(k-1)<=n

若x>m 取x=m 然后重新判断是否可以到 i 组

 

//#include <bits/stdc++.h>
#include<iostream>
#include<stack>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
typedef long long ll;
#define inf 2147483647
const ll INF = 0x3f3f3f3f3f3f3f3fll;
#define ri register int
template <class T> inline T min(T a, T b, T c)
{
    return min(min(a, b), c);
}
template <class T> inline T max(T a, T b, T c)
{
    return max(max(a, b), c);
}
template <class T> inline T min(T a, T b, T c, T d)
{
    return min(min(a, b), min(c, d));
}
template <class T> inline T max(T a, T b, T c, T d)
{
    return max(max(a, b), max(c, d));
}
#define scanf1(x) scanf("%d", &x)
#define scanf2(x, y) scanf("%d%d", &x, &y)
#define scanf3(x, y, z) scanf("%d%d%d", &x, &y, &z)
#define scanf4(x, y, z, X) scanf("%d%d%d%d", &x, &y, &z, &X)
#define pi acos(-1)
#define me(x, y) memset(x, y, sizeof(x));
#define For(i, a, b) for (ll i = (a); i <= (b); i++)
#define FFor(i, a, b) for (int i = a; i >= (b); i--)
#define bug printf("***********\n");
#define mp make_pair
#define pb push_back
const int N = 2000;
const int M=200005;
// name*******************************
ll n,k,m,d;
ll t;
ll ans=0;
// function******************************

//***************************************
int main()
{
    //    ios::sync_with_stdio(0);
    //    cin.tie(0);
    //     freopen("test.txt", "r", stdin);
    //  freopen("outout.txt","w",stdout);
    cin>>n>>k>>m>>d;
    For(i,1,d)
    {
        ll x=n/((i-1)*k+1);
        if(!x)break;
        if(x>m)x=m;
        if(n/(k*x)+((n%(k*x)>=x)?1:0)!=i)continue;
        ans=max(ans,x*i);
    }
    cout<<ans;
    return 0;
}