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Home Code Solutions Hackerrank Algorithms

Accessory Collection – HackerRank Solution

Accessory Collection - HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

bhautik bhalala by bhautik bhalala
May 27, 2022
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Table of Contents

  • Accessory Collection – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • Solutions of Algorithms Data Structures Hard HackerRank:
    • Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts
  • C++ Accessory Collection HackerRank Solution
  • Java Accessory Collection HackerRank Solution
  • Python 3 Accessory Collection HackerRank Solution
  • Python 2 Accessory Collection HackerRank Solution
  • C Accessory Collection HackerRank Solution
    • Warmup Implementation Strings Sorting Search Graph Theory Greedy Dynamic Programming Constructive Algorithms Bit Manipulation Recursion Game Theory NP Complete Debugging
    • Leave a comment below
      • Related posts:

Accessory Collection – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Solutions of Algorithms Data Structures Hard HackerRank:

Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts

C++ Accessory Collection HackerRank Solution


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#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<bitset>
#include<utility>
#include<functional>
#include<iomanip>
#include<sstream>
#include<ctime>
#include<cassert>
using namespace std;
#define y0 y0z
#define y1 y1z
#define yn ynz
#define j0 j0z
#define j1 j1z
#define jn jnz
#define tm tmz
#define buli(x) (__builtin_popcountll(x))
#define bur0(x) (__builtin_ctzll(x))
#define bul2(x) (63-__builtin_clzll(x))
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fil(a,b) memset((a),(b),sizeof(a))
#define cl(a) fil(a,0)
#define siz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define rep(i,a,b) for (int i=(a),_ed=(b);i<_ed;i++)
#define per(i,a,b) for (int i=(b)-1,_ed=(a);i>=_ed;i--)
#define pw(x) ((ll(1))<<(x))
#define upmo(a,b) (((a)=((a)+(b))%mo)<0?(a)+=mo:(a))
#define mmo(a,b) (((a)=1ll*(a)*(b)%mo)<0?(a)+=mo:(a))
void getre(){int x=0;printf("%d\n",1/x);}
void gettle(){int res=1;while(1)res<<=1;printf("%d\n",res);}
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<pii> vpii;
template<typename T,typename S>inline bool upmin(T&a,const S&b){return a>b?a=b,1:0;}
template<typename T,typename S>inline bool upmax(T&a,const S&b){return a<b?a=b,1:0;}
template<typename N,typename PN>inline N flo(N a,PN b){return a>=0?a/b:-((-a-1)/b)-1;}
template<typename N,typename PN>inline N cei(N a,PN b){return a>0?(a-1)/b+1:-(-a/b);}
template<typename N>N gcd(N a,N b){return b?gcd(b,a%b):a;}
template<typename N>inline int sgn(N a){return a>0?1:(a<0?-1:0);}
#if ( ( _WIN32 || __WIN32__ ) && __cplusplus < 201103L)
    #define lld "%I64d"
#else
    #define lld "%lld"
#endif
inline void gn(long long&x){
	int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
	while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;
}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
inline void gn(double&x){double t;scanf("%lf",&t);x=t;}
inline void gn(long double&x){double t;scanf("%lf",&t);x=t;}
inline void gs(char *s){scanf("%s",s);}
inline void gc(char &c){while((c=getchar())>126 || c<33);}
inline void pc(char c){putchar(c);}
#ifdef JCVB
#define debug(...) fprintf(stderr, __VA_ARGS__)
#else
#define debug(...) 
#endif
typedef long long ll;
typedef double db;
inline ll sqr(ll a){return a*a;}
inline db sqrf(db a){return a*a;}
const int inf=0x3f3f3f3f;
const db pi=3.14159265358979323846264338327950288L;
const db eps=1e-6;
//const int mo=0;
//int qp(int a,ll b){int n=1;do{if(b&1)n=1ll*n*a%mo;a=1ll*a*a%mo;}while(b>>=1);return n;}

int d,n,l,a;


ll sum(int l,int r){
	return 1ll*(l+r)*(r-l+1)/2;
}
inline ll val(int x){
	int lef=l-(n-1);
	int num=lef/x;
	int res=lef%x;
	ll ans=sum(a-d+1-num+1,a-d+1)*x;
	ans+=(a-d+1ll-num)*res;

	ans+=sum(a-d+2,a-1)*x;
	res=n-1-x*(d-2);
	ans+=1ll*res*a;
	return ans;
}
int main()
{
#ifdef JCVB
	//freopen("1.in","r",stdin);
	//freopen("1.out","w",stdout);
	int _time_jc=clock();
#endif
	int te;gn(te);
	while(te--){
		gn(l);gn(a);gn(n);gn(d);

		if(d==1){
			printf(lld"\n",1ll*a*l);
			continue;
		}
		if(a<d){
			printf("SAD\n");
			continue;
		}
		int lef=l-(n-1);

		int l1=cei(lef,a-d+1);
		int r1=flo(n-1,d-1);
		ll ma=0;
		if(l1>r1){
			printf("SAD\n");
			continue;
		}
		for (int x=l1;x<=r1;x++)upmax(ma,val(x));
		//printf(lld"\n",max(val(l1),val(r1)));
		printf(lld"\n",ma);
	}
#ifdef JCVB
	debug("time: %d\n",int(clock()-_time_jc));
#endif
	return 0;
}


Java Accessory Collection HackerRank Solution


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

public class Vic {

     static long Solve(int L, int A, int N, int D) {           

            if(D<2 || N>L)  // no constraint; use only element A
            	 return (long)L*A;
            else if(A<D || D>N) return -1; // no solution
            else { 
            	// a solution is always 
            	// bestsum of the best set of L elements where each subset 
              long bestsum = 0;
              for(int i = (int)Math.ceil((L-N+1.0)/(A-D+1.0)); i <= (N-1)/(D-1); i++){
            	int used = N-i*(D-2)-1; // number of A; used <= L.
            	long sum = (long)used*A;
            	if(D==2 && used>i){i=used;};//since i has not occurred
            	// the next num values A-1..A-num are taken min times
            	long num = (L-used)/i; //integer division will round down
            	sum += (num*i*(2*A-1-num))/2; 
            	used += num*i;
            	// a last value keeps the rest
            	sum += (L-used)*(A-num-1);
            	if(sum>bestsum) bestsum=sum;
              }
              return bestsum;
            }          	       		           	
        } //Solve

    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();
            int A = in.nextInt();
            int N = in.nextInt();
            int D = in.nextInt();
            long res = Solve(L,A,N,D);
            if(res<=0)
            	System.out.println("SAD");
            else
            	System.out.println(res);
        } // for
    } //main
    
}



Python 3 Accessory Collection HackerRank Solution


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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')



Python 2 Accessory Collection HackerRank Solution


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#!/bin/python
for cas in xrange(input()):
    l,a,n,d = map(int, raw_input().strip().split())
    if d == 1:
        res = l*a
    else:
        res = 0
        for rep in xrange(1,l+1):
            m = n-1-rep*(d-1)
            if m < 0: break
            v = max(0,a-((l-m)/rep))
            ln = (a-v)*rep + m
            if ln < l and v == 0: continue
            total = (a-v)*(a-v+1)/2 * rep - (a-v)*(l-m) + a*l
            res = max(res, total)

    print res or 'SAD'



C Accessory Collection HackerRank Solution


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#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;
}

 

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