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) {
printf("SAD\n");
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){
System.out.println("SAD");
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;
}
System.out.println(max==0?"SAD":max);
}
}
}
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:
print('SAD')
continue
if A == 1:
print('SAD')
continue
if A == 2:
if D == 2 and (N - D + 1) * 2 >= L:
print(N - D + L + 1)
continue
else:
print('SAD')
continue
g = math.ceil((L - A - N + D) / (A - 2))
if g > N - D:
print('SAD')
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))
print(max_haul if max_haul else 'SAD')
View More Similar Problems
Left Rotation
A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d
View Solution →Sparse Arrays
There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun
View Solution →Array Manipulation
Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu
View Solution →Print the Elements of a Linked List
This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode
View Solution →Insert a Node at the Tail of a Linked List
You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink
View Solution →Insert a Node at the head of a Linked List
Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below
View Solution →
