 # BURGER - Editorial

Setter: Sunidhi Chandra
Tester: Istvan Nagy
Editorialist: Taranpreet Singh

Easy-Medium

Greedy

# PROBLEM

You have taken an eating challenge from Chef, to eat exactly Y burgers. You will eat in the following way.

• In the first minute, you eat exactly X burgers, and in every minute after that, you will eat exactly twice the number of burgers you ate in the previous minute.
• You can take a break of exactly one minute.
• When you start eating again after the break, your eating streak resets, and you eat X burgers in that minute, and so on.

Additionally, let a_1, a_2 \ldots a_k denote the length of your eating streaks. Chef asks that all a_i are pairwise distinct.

Find the minimum number of minutes you need in order to eat exactly Y burgers or determine if it’s impossible to do so.

# QUICK EXPLANATION

• A streak of d minutes involves eating exactly X*(2^{d+1}-1) burgers, and takes d minutes eating, and 1 minute break.
• We can make a set of streaks possible, since the number of burgers eaten in a streak rises exponentially, we don’t need to store streaks of length more than log_2(Y/X).
• Now, considering these streaks in descending order of the number of burgers, while the number of burgers to be eaten is greater or equal to the number of burgers in a streak, it is optimal to include that streak in the set and update the number of burgers to be eaten.
• If, in the end, all burgers are not eaten, it is impossible to eat them all in this setting.

# EXPLANATION

Let us notice first that if X doesn’t divide Y, it’s not possible to eat Y burgers at all, since every minute, we eat burgers in multiples of X. So, if X doesn’t divide Y, it’s an impossible case.

Now assuming Y = T*X, so we need to eat T burgers if our streak starts with exactly 1 burger.

Let us see the number of burgers eaten in a streak of d days. We can see that number of burgers eaten are \displaystyle \sum_{i = 0}^{d-1} 2^i, which is a geometric progression, so the number of burgers eaten shall be 2^d-1.

Let us assume, in a streak, the last day, we take a break. Now, the number of burgers eaten in a streak of d days is 2^{d-1}-1. Also, d \geq 2, since in each streak, we have at least one day we eat, and the last day is spent on a break. This way, the minute spent on break is already included, except for the last streak.

### Redefined Goal

Our goal is to find a set A of pairwise integers greater than 1 such that \displaystyle \sum_{x \in A} (2^{x-1}-1) = T (The number of burgers eaten) and \displaystyle -1 + \sum_{x \in A} x is minimized (Sum of length of streaks). One is subtracted since the last streak also includes one minute for a break, which can be skipped.

Since 2^{d-1}-1 \leq T for d \in A, So we have d-1 \leq log_2(T+1). Hence, we only need to consider positive integers d up to log_2(T+1)+1

Hence, considering first streaks of length up to log_2(T+1), we want to check if there’s a subset with burger count T, and if multiple subsets exist, find the one with minimum sum. The subset cannot contain duplicates.

Let’s consider the streaks now, their duration, and the number of burgers eaten

duration               burgers eaten
2                                 1
3                                 3
4                                 7
5                                15
6                                31
...


### Why greedy selection work

Claim: Considering streaks in descending order, if 2^{d-1}-1 \leq T holds for current streak d, then if a valid subset exists, it shall include a streak of length d.

Proof

The number of burgers eaten in streak of length d is 2^{d-1}-1. Let’s assume current streak is not included. So the remaining streak must eat T burgers. The sum of number of burgers in remaining streaks is given by \displaystyle \sum_{x = 2}^{d-1} (2^{x-1}-1) = \sum_{x = 1}^{d-2} 2^x -(d-2). By GP sum, we have \displaystyle \sum_{x = 1}^{d-2} 2^x = 2^{d-1}-2.

Hence, by taking all remaining streaks, we manage to eat 2^{d-1}-d burgers.

But 2^{d-1}-d < 2^d-1 \leq T, so by skipping streak of length d, we can never eat all burgers while keeping the remaining streaks pairwise distinct.

Hence, if a streak of length d is possible, we must include it in our set of streaks.

The above claim allows us to greedily consider streaks from log_2(T) to 2 in decreasing order, check if we can include that streak in our subset. If we can, we must, and we update the number of burgers to be eaten. If at the end of the process, all burgers are not eaten, it means it is impossible to eat them all.

However, if we managed to eat streaks, one less than the sum of lengths of streaks shall give the minimum number of minutes, thus, solving the problem.

We can make the list of streaks beforehand. Refer to my implementation in the case facing any issues.

### Bonus

Prove that the minimum time taken to eat all burgers, if possible, is also the same as the maximum time possible under the problem constraints.

# TIME COMPLEXITY

The time complexity is O(log_2(Y/X)) per test case.

# SOLUTIONS

Setter's Solution
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define fastIO ios_base::sync_with_stdio(false); cin.tie(0)

int main()
{
fastIO;
int t=1;
cin>>t;
while(t--){
ll x,y,ans=-1;
cin>>x>>y;
if((y%x)==0){
y/=x;
for(int k=1;k<60;k++){
ll val=y+k;
if(val%2)continue;
int cnt=__builtin_popcountll(val);
if(cnt==k){
int tmp=0;
for(int i=1;i<=60;i++){
if(val&(1LL<<i))tmp+=i;
}
tmp+=k-1;
ans=tmp;
break;
}
}
}
cout<<ans<<endl;
}

return 0;
}

Tester's Solution
#include <iostream>
#include <cassert>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <random>
#include <limits>
#include <functional>

#ifdef HOME
#define NOMINMAX
#include <windows.h>
#endif

#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define forn(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)

template<class T> bool umin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool umax(T &a, T b) { return a < b ? (a = b, true) : false; }

using namespace std;

long long readInt(long long l, long long r, char endd) {
long long x = 0;
int cnt = 0;
int fi = -1;
bool is_neg = false;
while (true) {
char g = getchar();
if (g == '-') {
assert(fi == -1);
is_neg = true;
continue;
}
if ('0' <= g && g <= '9') {
x *= 10;
x += g - '0';
if (cnt == 0) {
fi = g - '0';
}
cnt++;
assert(fi != 0 || cnt == 1);
assert(fi != 0 || is_neg == false);

assert(!(cnt > 19 || (cnt == 19 && fi > 1)));
}
else if (g == endd) {
assert(cnt > 0);
if (is_neg) {
x = -x;
}
assert(l <= x && x <= r);
return x;
}
else {
assert(false);
}
}
}

string readString(int l, int r, char endd) {
string ret = "";
int cnt = 0;
while (true) {
char g = getchar();
assert(g != -1);
if (g == endd) {
break;
}
cnt++;
ret += g;
}
assert(l <= cnt && cnt <= r);
return ret;
}
long long readIntSp(long long l, long long r) {
}
long long readIntLn(long long l, long long r) {
}
string readStringLn(int l, int r) {
}
string readStringSp(int l, int r) {
}

uint32_t lzCount(uint64_t v)
{
#ifdef WIN32
return static_cast<uint32_t>(__lzcnt64(v));
#else
return __builtin_clzll(v);
#endif
}

int main(int argc, char** argv)
{
#ifdef HOME
if(IsDebuggerPresent())
{
freopen("../in.txt", "rb", stdin);
freopen("../out.txt", "wb", stdout);
}
#endif

forn(tc, T)
{
if (Y % X)
{
printf("-1\n");
continue;
}
Y /= X;
int64_t lv = numeric_limits<int64_t>::max();
int32_t res = 0;
while (Y)
{
uint32_t lzc = lzCount(Y);
int64_t val = (1ll << (64 - lzc)) - 1;
if (val > Y)
{
val = (1ll << (63 - lzc)) - 1;
res += 64 - lzc;
}
else
res += 65 - lzc;

if(lv == val)
{
res = 0;
break;
}
Y -= val;
lv = val;
}
--res;
printf("%d\n", res);
}
assert(getchar() == -1);
return 0;
}

Editorialist's Solution
import java.util.*;
import java.io.*;
class BURGER{
//SOLUTION BEGIN
int MX = 62;
long[] burgerCount, duration;
void pre() throws Exception{
burgerCount = new long[MX];
duration = new long[MX];

burgerCount = 1;
duration = 2;
for(int i = 1; i< MX; i++){
burgerCount[i] = burgerCount[i-1]*2+1;
duration[i] = duration[i-1]+1;
}
}
void solve(int TC) throws Exception{
long X = nl(), Y = nl();
if(Y%X != 0){
pn(-1);
return;
}
long TotalBurgers = Y/X;
long days = 0;
for(int i = MX-1; i>= 0; i--){
if(TotalBurgers >= burgerCount[i]){
TotalBurgers -= burgerCount[i];
days += duration[i];
}
}
if(TotalBurgers == 0)pn(days-1);
else pn(-1);
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static boolean multipleTC = true;
void run() throws Exception{
//        out = new PrintWriter("out.txt");
out = new PrintWriter(System.out);
//Solution Credits: Taranpreet Singh
int T = (multipleTC)?ni():1;
pre();for(int t = 1; t<= T; t++)solve(t);
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
new BURGER().run();
}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object o){out.print(o);}
void pn(Object o){out.println(o);}
void pni(Object o){out.println(o);out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}

StringTokenizer st;
}

}

String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
}catch (IOException  e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}

String nextLine() throws Exception{
String str = "";
try{
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
}


Feel free to share your approach. Suggestions are welcomed as always. 2 Likes

Can someone please provide a test case where this fails. I have used the same idea, but maybe I am missing something. I have tried many cases but can’t come up with something.
https://www.codechef.com/viewsolution/46832400
Thanks!

Can someone explain the testcase
1
5 10
According to my solution, answer should be 3
But the editorial solution is giving -1

Can anyone expain this testcase please

1 Like

Yes my solution also gives 3 for it , while editorial’s solution gives -1 . Did I read the question wrong?

1 Like

It should give -1. The reason being all streaks have to be unique
For this to give 3 ur logic suggests:
5 0 5
But 1st 5 and 2nd 5 are same streak so -1 should be returned.

5 Likes

oh got it , thanks !

1 Like

I dont even think of that case F
Thanks for the clarification

2 Likes

@taran_1407 Can you please tell why this solution is wrong?

https://www.codechef.com/viewsolution/46834287
can anyone give me a case for which my code gives Wrong answer?

You are using int. Read the constraints.
PS - avoid using builtin pow.

how is the complexity log(y/x) isn’t it supposed to be log(y/x)^2 ?

For every i from 1 to 60 we are also using builtinpopcount whose complexity is the number of bits.

In the setter’s solution

@taran_1407

Hi Codechef admin, I have a small doubt about why do we do Y/X before starting the for loop for greedy approach ?

My code was not getting accepted untill i manually calculated number of one’s(set bits) in a number and not using builtin popcount function can any body tell the reason behind it.
Accepted code
Rejected code

1 Like

Can anyone tell me why or on what test cases is this solution wrong.
https://www.codechef.com/viewsolution/46839451

Hey, I found few test cases. Unfortunately, the values are so large. Here are few test cases where your code fails:

1 9189666384
1 8887866239
1 4827902913
1 8396752774


It took me 30 mins to find what was going wrong. If this doesn’t help - 1 Like

Same. Huge numbers.

1 9520884497965793
31 9769873759224593
1 9365680313404807


That’s because __builtin_popcount accepts a parameter as unsigned int x, for large numbers this will overflow and give wrong results. For large numbers there are 2 more inbuilt functions namely __builtin_popcountl (accepts unsigned long) and __builtin_popcountll (accepts unsigned long long), use the long long variant here and you will get AC.

2 Likes

I changed the built in power function to pre computed values of 2, these testcases passed, but still it is giving WA.
https://www.codechef.com/viewsolution/46840413

I referred to my solution while mentioning time complexity.

In my solution, I’m not using bitcount. Just iterating over log(Y/X) streak lengths in descending order.

Right I did mention setter’s solution though. Was it intended to allow those solutions to pass?