# Accessory Collection

### Problem Statement :

```Victoria is splurging on expensive accessories at her favorite stores. Each store stocks  types of accessories, where the  accessory costs  dollars (). Assume that an item's type identifier is the same as its cost, and the store has an unlimited supply of each accessory.

Victoria wants to purchase a total of  accessories according to the following rule:

For example, if , , and , then she must choose  accessories such that any subset of  of the  accessories will contain at least  distinct types of items.

Given , , , and  values for  shopping trips, find and print the maximum amount of money that Victoria can spend during each trip; if it's not possible for Victoria to make a purchase during a certain trip, print SAD instead. You must print your answer for each trip on a new line.

Input Format

The first line contains an integer, , denoting the number of shopping trips.
Each of the  subsequent lines describes a single shopping trip as four space-separated integers corresponding to L, A , N , and D, respectively.

Output Format

For each shopping trip, print a single line containing either the maximum amount of money Victoria can spend; if there is no collection of items satisfying her shopping rule for the trip's L, A , N , and D  values, print SAD instead.```

### Solution :

```                            ```Solution in C :

In   C  :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
int T;
scanf("%d",&T);
for(int a0 = 0; a0 < T; a0++){
int L;
int A;
int N;
int D;
scanf("%d %d %d %d",&L,&A,&N,&D);
if (D == 1) {
printf("%lld\n", (long long)A*L);
continue;
}
int max, min, i, q, r;
long long count, result = 0;
max = (N-1)/(D-1);
if ((L-(N-1)+max-1)/max+D-1 > A) {
continue;
}
min = (L-(N-1)+A-(D-1)-1)/(A-(D-1));
for (i=max; i>=min; i--) {
q = (L-(N-1))/i;
r = (L-(N-1))%i;
count = (long long)(A-(D-1+q))*r + (long long)(A-(D-1+q)+1+A-1)*(D-2+q)/2*i + (long long)A*(N-1-i*(D-2));
if (count <= result) {
break;
}
result = count;
}
printf("%lld\n", result);
}
return 0;
}```
```

```                        ```Solution in C++ :

In  C ++  :

#include <bits/stdc++.h>
using namespace std;
#define fo(i,a,b) for(int i=(a);i<(b);i++)
#define MOD 1000000007
#define MT make_tuple
#define PB push_back
typedef long long ll;

int tst;
ll T, A, N, D, mx;

ll g (ll i, ll j) {
return j*(j+1)/2 - i*(i-1)/2;
}

int main () {
cin >> tst;
while (tst--) {
mx = -1;
cin >> T >> A >> N >> D;
if (D == 1) {
cout << T*A << '\n';
continue;
}
for (ll l = 0; l <= T; l++) {
ll sum = 0;
ll x = T-N+1, y = N-1;
ll ls = A-(D-1), rs = D-1;
if (ls * l < x) continue;
if (rs * l > y) continue;

ll lw = x / l; //this is how many full ones on left
ll lrem = x - lw * l; //this is how many remain
//from A-D+1-lw+1 to A-D+1, times l
sum += g(A-D+1-lw+1, A-D+1) * l + lrem * (A-D+1-lw);
sum += g(A-D+2, A-1) * l + (y - l * (D-2)) * A;
mx = max(mx, sum);
}
if (mx == -1) cout << "SAD" << '\n';
else cout << mx << '\n';
}
return 0;
}```
```

```                        ```Solution in Java :

In   Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
for(int a0 = 0; a0 < T; a0++){
int L = in.nextInt();
long A = in.nextInt();
int N = in.nextInt();
int D = in.nextInt();

if (N<D || N>L || A<D){
continue;
}

// Deal with the special case
if (D==1){
System.out.println(L*A);
continue;
}

// The number of accessories
// A : a1
// A-1 to A-n: a2
// A-n-1: a3
long max = 0;
int a2Max = (N-1)/(D-1);
// Loop start from maximun a2
for (int a2=a2Max;a2>=1; a2--){
// Calculate a1, a3, and n by a2
long a1 = N + (a2-1) - a2*(D-1);
long n = (L-a1)/a2;
long a3 = (L-a1)%a2;
// Break when the type of accessories (A) is not enough
if (n>A-1 || (n==A-1 && a3 > 0)){
break;
}
// Caclulate cost
long sum = A*a1 + (A-1+A-n)*n/2*a2 + a3 * (A-n-1);
// Break when cost starts decreasing
if (sum<=max){
break;
}
max = sum;
}
}
}
}```
```

```                        ```Solution in Python :

In  Python3 :

import math

def total(g):
# g = size of consecutive groups after the first one
if g <= 0:
return 0
first = N - g * (D - 2) - 1
f = (L - first) // g
if first < g or A - f <= 0 or first + g * (A - 1) < L:
return 0
partial = int((2 * (A - 1) - f + 1) / 2 * f)
left = (A - f - 1) * (-f * g + L - first)
return first * A + partial * g + left

for _ in range(int(input())):
L, A, N, D = map(int, input().strip().split())
if D == 1:
print(L * A)
continue
if D == 2:
n1 = (N - 1)
if n1 * A >= L:
ln1 = L // n1
print(int((2 * A + 1 - ln1) / 2 * ln1 * n1) + (A - ln1) * (L - ln1 * n1))
continue
else:
continue
if A == 1:
continue
if A == 2:
if D == 2 and (N - D + 1) * 2 >= L:
print(N - D + L + 1)
continue
else:
continue

g = math.ceil((L - A - N + D) / (A - 2))
if g > N - D:
continue
else:
# I did the math
# ArgMax of total, A > 2 and D > 2
argmax = int(math.sqrt((2 - 3 * D + D * D) * (1 + L - N) * (1 + L - N)) / (2 - 3 * D + D * D)) + 1
if (D == 3 or D == 4 and A > 2) or (D > 4 and A > D - 2):
# region where enough items are in A (A - f > 0)
argmax = max(int((1 + L - N) / (2 + A - D)), argmax)
# region where first group is the largest (first >= g)
argmax = min(int((N - 1) / (D - 1)), argmax)
# region where total number of items is at least L (first + g * (A - 1) >= L)
if D == 3 or A > D - 1:
argmax = max(int((1 + L - N) / (1 + A - D)), argmax)
elif D > 3 and A < D - 1:
argmax = min(int((1 + L - N) / (1 + A - D)), argmax)
max_haul = max(total(argmax - 1), total(argmax), total(argmax + 1))
```

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