Bytelandian gold coins
Problem Statement :
In Byteland they have a very strange monetary system. Each Bytelandian gold coin has an integer number written on it. A coin n can be exchanged in a bank into three coins: n/2, n/3 and n/4. But these numbers are all rounded down (the banks have to make a profit). You can also sell Bytelandian coins for American dollars. The exchange rate is 1:1. But you can not buy Bytelandian coins. You have one gold coin. What is the maximum amount of American dollars you can get for it? Input The input will contain several test cases (not more than 10). Each testcase is a single line with a number n, 0 <= n <= 1 000 000 000. It is the number written on your coin. Output For each test case output a single line, containing the maximum amount of American dollars you can make. Example Input: 12 2 Output: 13 2 You can change 12 into 6, 4 and 3, and then change these into 6+4+3=13. If you try changing the coin 2 into 3 smaller coins, you will get 1, 0 and 0, and later you can get no more than 1 out of them. It is better just to change the 2 coin directly into 2.
Solution :
Solution in C :
#include<stdio.h>
long long int dp[10000000];
void initialize()
{
int i;
for(i=1;i<=10000000;i++)
dp[i]=0;
}
long long int coins(long long int n)
{
long long int ans;
if(n<12)
return n;
if(n<10000000 && dp[n]!=0)
{
return dp[n];
}
ans=coins(n/2)+coins(n/3)+coins(n/4);
if(n<10000000)
{
ans=coins(n/2)+coins(n/3)+coins(n/4);
dp[n]=ans;
}
return ans;
}
int main()
{
long long int n,res;
initialize();
while(scanf("%lld",&n)!=EOF)
{
res=coins(n);
printf("%lld\n",res);
}
}
Solution in C++ :
#include <bits/stdc++.h>
#define inf INT_MAX
#define mod 1e9+7
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
typedef long long ll;
#define pb push_back
#define lcm(a,b) (a*b)/(__gcd(a,b))
#define fw(i,a,b) for(int i=a;i<b;i++)
#define vl vector<ll>
#define vvl vector<vl>
using namespace std;
ll power(ll a , ll b,ll modi)
{
a%=modi;
ll res = 1 ;
while(b)
{
if(b%2) {
res = (res * a) % modi ;
}
b/=2 ;
a = (a*a) % modi ;
}
return res ;
}
map<ll , ll > dp;
ll Dp(ll n)
{
if(dp.find(n)==dp.end())dp[n] = max(Dp(n/2) + Dp(n/3) + Dp(n/4), n);
return dp[n];
}
ll fun(ll n)
{
if(n<12)return dp[n]=n;
if(dp[n]) return dp[n];
return dp[n]=max(n,fun(n/2)+fun(n/3)+fun(n/4));
}
int main()
{
ll n;
while(cin>>n)
{
cout<<fun(n)<<endl;
}
return 0;
}
Solution in Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
import java.lang.*;
public class Main{
public static PrintWriter out;
public static FastReader in;
public static int mod = 1000003;
static class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br=new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while(st==null || !st.hasMoreTokens()){
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String str="";
try {
str=br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
try {
in = new FastReader();
out = new PrintWriter(new BufferedOutputStream(System.out));
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
long n = sc.nextLong();
long ans = solve(n);
out.println(ans);
}
out.flush();
out.close();
} catch (Exception e) {
System.out.println("Exception");
return;
}
}
public static long solve(long n ){
long res = n/2+n/3+n/4;
if(n<res){
return solve(n/2)+solve(n/3)+solve(n/4);
}else{
return n;
}
}
/*
// Points to rememeber:
- if a question is asked where you are given two condition one where you have to include
and one in which you have to exclude them use dp memoization (maximum in non-adjecent array)
- In a problem which has distance given by two point use graph and if we need to find the
element in between sort in the pair array for values instead of keys
- Use pascal's triangle to solve nCr problem.
- b^-1%m = b^m-2 (Fermat's little theorem). (nCr problem)
- if you have a^10 (use fast exponenetiation or fast factorial computation)
- xor adds if one of the two number are power of 2 and the other is less than it and
it subtracts if it is more. (8^7 == 15 but 8^9 == 1).
- even number with xor 1 adds and odd subtracts.
//String:
- total possible substring -> n * (n+1)/2, total permutations -> n!
- You can use Manacher’s algorithm to find and longest palindromic substirng
and also modify it to find count pallindromic substrings.
- you can also use dp (diagonal method tabulation) to count pallindromic substrings.
*/
static int highestPowerOf2(int x)
{
// check for the set bits
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
// Then we remove all but the top bit by xor'ing the
// string of 1's with that string of 1's shifted one to
// the left, and we end up with just the one top bit
// followed by 0's.
return x ^ (x >> 1);
}
static boolean isPrime(int a){
if (a == 1) return false;
if (a == 2 || a == 3) return true;
if (a%2==0 || a%3==0) return false;
for (int i=5;i*i<=a;i+=6){
if (a%i==0 || a%(i+2)==0) return false;
}
return true;
}
static void printAllPrime(int n){
// Sieve of Eratosthenes algorithm
if(n <= 1) return;
boolean[] arr = new boolean[n+1];
Arrays.fill(arr,true);
for (int i=2;i*i<=n;i++){
if (arr[i]){
for (int j=i*i;j<=n;j+=i){
arr[j] = false;
}
}
}
for (int i=2;i<=n;i++){
if (arr[i]){
System.out.printf(i+" ");
}
}
}
static long pow(long a,long b){
long ans = b;
long res =1;
if(b<0){
ans = -1*b;
}
while(ans>0){
if((ans&1)==1){
res = (res*a)%mod;
}
a = (a*a)%mod;
ans = ans>>1;
}
if(b<0){
res = 1/res;
}
return res;
}
static double pow(double a,long b){
long ans = b;
double res =1;
if(b<0){
ans = -1*b;
}
while(ans>0){
if((ans&1)==1){
res = (res*a)%mod;
}
a = (a*a)%mod;
ans = ans>>1;
}
if(b<0){
res = 1/res;
}
return res;
}
static void sort(int[] arr)
{
ArrayList<Integer> ls = new ArrayList<Integer>();
for(int x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
static void sort(long[] arr)
{
ArrayList<Long> ls = new ArrayList<Long>();
for(long x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
static HashMap<Integer, Integer> sortValue(HashMap<Integer,Integer> h){
List<Map.Entry<Integer, Integer> > list = new ArrayList<>(h.entrySet());
// Sort the list
Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() {
public int compare(Map.Entry<Integer, Integer> o1, Map.Entry<Integer, Integer> o2){
int fp = o2.getValue().compareTo(o1.getValue());
int sp = o2.getKey().compareTo(o1.getKey());
return fp==0 ? sp : fp;
}
});
//clear the hashmap
// h.clear();
HashMap<Integer, Integer> hm = new LinkedHashMap<>();
for(Map.Entry<Integer, Integer> mp : list){
hm.put(mp.getKey(),mp.getValue());
}
return hm;
}
static HashMap<Integer, Integer> sortKey(HashMap<Integer,Integer> h){
List<Map.Entry<Integer, Integer> > list = new ArrayList<>(h.entrySet());
// Sort the list
Collections.sort(list, new Comparator<Map.Entry<Integer, Integer> >() {
public int compare(Map.Entry<Integer, Integer> o1, Map.Entry<Integer, Integer> o2){
int fp = o2.getValue().compareTo(o1.getValue());
int sp = o2.getKey().compareTo(o1.getKey());
return fp==0 ? fp : sp;
}
});
//clear the hashmap
// h.clear();
HashMap<Integer, Integer> hm = new LinkedHashMap<>();
for(Map.Entry<Integer, Integer> mp : list){
hm.put(mp.getKey(),mp.getValue());
}
return hm;
}
static long totient(long n)
{
long result = n;
for (int p = 2; p*p <= n; ++p)
if (n % p == 0)
{
while(n%p == 0)
n /= p;
result -= result/p;
}
if (n > 1)
result -= result/n;
return result;
}
static int pow(int x,int y){
return (int)Math.pow(x,y);
}
static void allDivisor(int a){
int i=0;
for (i=1;i*i<=a;i++){
if (a%i==0){
System.out.printf(i+" ");
}
}
for (;i>=1;i--){
if (a%i==0){
System.out.printf(a/i+" ");
}
}
}
static int binaryToInt(String s){
return Integer.parseInt(s,2);
}
static String toBinaryString(int s){
return Integer.toBinaryString(s);
}
static void primeFactors(int a){
if (a<=1) return;
while (a%2==0) {System.out.printf(2+" "); a=a/2;}
while (a%3==0) {System.out.printf(3+" "); a=a/3;}
for (int i=5;i*i<=a;i+=6){
while (a%i==0){
System.out.printf(i+" ");
a = a/i;
}
while (a%(i+2)==0){
System.out.printf((i+2)+" ");
a = a / (i+2);
}
}
if (a>3){
System.out.printf(a+" ");
}
System.out.println();
}
static int lcm(int a,int b){
return a*b/gcd(a,b);
}
static int gcd(int a, int b){
if (a==0) return b;
return gcd(b%a,a);
}
static int countZeros(int f){
int x = 0;
for (int i=5;i<=f;i=i*5){
x += f/i;
}
return x;
}
static int ExtendedGcd(int a, int b,int x,int y){
if(b == 0){
x = 1;
y = 0;
return a;
}
int x1=1,y1=1;
int ans = ExtendedGcd(b, a%b,x1,y1);
x = y1;
y = x1 - (a/b)*y1;
System.out.println(x+" "+y);
return ans;
}
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////Lazy work/////////////////////////////////////////////////////////////
static void input(int[] a){
for (int i=0;i<a.length;i++){
a[i] = in.nextInt();
}
}
static void input(long[] a){
for (int i=0;i<a.length;i++){
a[i] = in.nextLong();
}
}
static void input(String[] a){
for (int i=0;i<a.length;i++){
a[i] = in.next();
}
}
static void swap(char[] c,int i,int j){
char t = c[i];
c[i] = c[j];
c[j] = t;
}
static void swap(int[] c,int i,int j){
int t = c[i];
c[i] = c[j];
c[j] = t;
}
static void swap(long[] c,int i,int j){
long t = c[i];
c[i] = c[j];
c[j] = t;
}
static void print(int[] a){
for (int i=0;i<a.length;i++){
out.printf(a[i]+" ");
}
out.println();
}
static void print(int[][] a){
for (int i=0;i<a.length;i++){
for(int j=0;j<a[i].length;j++){
out.printf(a[i][j]+" ");
}
out.println();
}
}
static void print(long[] a){
for (int i=0;i<a.length;i++){
out.printf(a[i]+" ");
}
out.println();
}
static void print(char[] a){
for (int i=0;i<a.length;i++){
out.printf(a[i]+" ");
}
out.println();
}
static void print(String s){
for (int i=0;i<s.length();i++){
out.printf(s.charAt(i)+" ");
}
out.println();
}
static void print(ArrayList<Integer> a){
a.forEach(e -> out.printf(e+" "));
out.println();
}
static void print(LinkedList<Integer> a){
a.forEach(e -> out.printf(e+" "));
out.println();
}
static void print(HashSet<Integer> a){
a.forEach(e -> out.printf(e+" "));
out.println();
}
static void print(HashMap<Integer,Integer> a){
for(Map.Entry<Integer, Integer> mp : a.entrySet()){
out.println(mp.getKey() + " "+ mp.getValue());
}
}
static void reverse(int[] a){
int i=0,j=a.length-1;
while(i<j){
int t = a[i];
a[i] = a[j];
a[j]= t;
i++;
j--;
}
}
static String reverse(String s){
char[] a = s.toCharArray();
int i=0,j=a.length-1;
while(i<j){
char t = a[i];
a[i] = a[j];
a[j]= t;
i++;
j--;
}
return String.valueOf(a);
}
}
class CompareP implements Comparator<Pair> {
public int compare(Pair a, Pair b){
long dif = a.v - b.v;
if (dif > 0) return 1;
if (dif < 0) return -1;
return 0;
}
}
class CompareT implements Comparator<Triplet> {
public int compare(Triplet a, Triplet b){
long dif = a.z - b.z;
if (dif > 0) return 1;
if (dif < 0) return -1;
return 0;
}
}
class Triplet{
long x;
long y;
long z;
public Triplet(long x,long y,long z){
this.x = x;
this.y = y;
this.z = z;
}
}
class Pair {
int k;
int v;
public Pair(int k, int v) {
this.k = k;
this.v = v;
}
}
class ncr
{
public int mod = 1000003;
public long[] fact = new long[mod+1];
public long[] ifact = new long[mod+1];
public int nCr(long n, long r)
{
preFactFermat();
long ans = 1;
// while(n>0 && r>0){
// int a=(int) (n%mod),b= (int)(r%mod);
// n = n/mod;r=r/mod;
// if(a<b){
// return 0;
// }else{
// ans = (ans* (fact[a]*((ifact[b]*ifact[a-b])%mod)%mod))%mod;
// }
// }
ans = lucas(n,r,ans);
return (int)ans;
}
public long lucas(long n,long r,long ans){
if(r==0)return 1;
long ni=n%mod,ri=r%mod;
return (lucas(n/mod,r/mod,ans)*(fermat(ni,ri,ans)%mod))%mod;
}
public long fermat(long n,long r,long ans){
if(n<r){
return 0;
}
ans = (ans* (fact[(int)n]*((ifact[(int)r]*ifact[(int)(n-r)])%mod)%mod))%mod;
return ans;
}
public void preFactFermat(){
fact[1] = 1;
fact[0] = 1;
ifact[0] = expo(1,mod-2);
ifact[1] = expo(1,mod-2);
for(int i=2;i<=mod;i++){
fact[i] = (i*(fact[i-1]%mod))%mod;
ifact[i] = expo(fact[i],mod-2);
}
}
public long expo(long a,long b){
long ans = b;
long res =1;
if(b<0){
ans = -1*b;
}
while(ans>0){
if((ans&1)==1){
res = (res*a)%mod;
}
a = (a*a)%mod;
ans = ans>>1;
}
if(b<0){
res = 1/res;
}
return res;
}
}
class FenwickTree{
int[] ft;
public void print(){
for (int i=0;i<ft.length;i++){
System.out.printf(ft[i]+" ");
}
}
public FenwickTree(int[] a){
ft = new int[a.length+1];
for (int i=0;i<a.length;i++){
this.update(i,a[i]);
}
}
public int getSum(int i){
int sum = 0;
while(i>0){
sum += ft[i];
i = i - (i & (-i));
}
return sum;
}
public void update(int i,int d){
i = i +1;
while(i<ft.length){
ft[i] += d;
i = i + (i &(-i));
}
}
}
class SegmentTree{
int[] st;
public SegmentTree(int[] a){
st = new int[a.length*4];
construct(a,0,a.length-1,0);
}
void print(){
for(int i=0;i<st.length;i++){
System.out.printf(st[i]+" ");
}
System.out.println();
}
int construct(int[] a,int ss,int se,int si){
if(ss==se){
st[si] = a[ss];
return a[ss];
}
int mid = (ss+se)/2;
st[si] = construct(a,ss,mid,2*si+1) + construct(a,mid+1,se,2*si+2);
return st[si];
}
int getSum(int qs,int qe,int ss,int se,int si){
if(qe<ss || qs>se){
return 0;
}
if(ss>=qs && se <= qe){
return st[si];
}
int mid = (ss+se)/2;
return getSum(qs,qe,ss,mid,2*si+1) + getSum(qs,qe,mid+1,se,2*si+2);
}
void update(int ss,int se,int si,int i,int diff){
if(ss > i || i> se){
return;
}
this.st[si] += diff;
if(ss< se){
int mid = (ss+se)/2;
update(ss,mid,2*si+1,i,diff);
update(mid+1,se,2*si+2,i,diff);
}
}
}
Solution in Python :
cache={i:i for i in range(12)}
def solve(i):
global cache
if i not in cache:
cache[i] = max(solve(i//2)+solve(i//3)+solve(i//4), i)
return cache[i]
while True:
try:
n=int(input())
except:break
print(solve(n))
View More Similar Problems
Median Updates
The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o
View Solution →Maximum Element
You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each
View Solution →Balanced Brackets
A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra
View Solution →Equal Stacks
ou have three stacks of cylinders where each cylinder has the same diameter, but they may vary in height. You can change the height of a stack by removing and discarding its topmost cylinder any number of times. Find the maximum possible height of the stacks such that all of the stacks are exactly the same height. This means you must remove zero or more cylinders from the top of zero or more of
View Solution →Game of Two Stacks
Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B = [b0, b1, . . . , b m-1] where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game: In each move, Nick can remove one integer from the top of either stack A or stack B. Nick keeps a running sum of the integers he removes from the two stacks. Nick is disqualified f
View Solution →Largest Rectangle
Skyline Real Estate Developers is planning to demolish a number of old, unoccupied buildings and construct a shopping mall in their place. Your task is to find the largest solid area in which the mall can be constructed. There are a number of buildings in a certain two-dimensional landscape. Each building has a height, given by . If you join adjacent buildings, they will form a solid rectangle
View Solution →