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Practice
Setter:
Testers: Tejas Pandey and Abhinav sharma
Editorialist: Taranpreet Singh
DIFFICULTY
Simple
PREREQUISITES
Math
PROBLEM
Chef is given a function F(X) such that the condition:
-
F(X) = F(2\cdot X)
holds true for all X, where X is a positive integer.
For a given positive integer N, find the maximum number of distinct elements in the array [F(1), F(2), F(3), \ldots, F(N)].
QUICK EXPLANATION
- F(X) will not contribute a distinct value in array for all even X since if F(X/2) is included in the array, then the same value is already included due to F(X/2).
- Hence, F(X) can contribute a distinct value only for odd values of X.
- The number of odd values in range [1, N] is \displaystyle\left\lceil\frac{N}{2}\right\rceil, which is the number of distinct values achievable.
EXPLANATION
Observation: F(X) will not contribute a distinct value in array for all even X.
Considering example X = 12. We know that F(6) = F(12). Hence, if N \geq 12, then N \geq 6, which means F(6) is also included in array. So the value F(12) is not a new value.
This implies that no even X can contribute a distinct value in array. We don’t have any constraint on odd values of X, so let’s find a function which contributes distinct values for each odd X, since that’s the best we can achieve here.
Let’s consider
F(n) =
\begin{cases}
F(n/2) & \quad \text{if } n \text{ is even}\\
n & \quad \text{if } n \text{ is odd}
\end{cases}
This function would contribute a distinct value for each odd X. This leads to \displaystyle\left\lceil\frac{N}{2}\right\rceil distinct values, which is maximum possible.
Hence, for given N, the number of distinct values achievable is \displaystyle\left\lceil\frac{N}{2}\right\rceil.
Bonus
- Prove \displaystyle\left\lceil\frac{N}{2}\right\rceil is equal to \displaystyle\left\lfloor\frac{N+1}{2}\right\rfloor.
- Prove \displaystyle\left\lceil\frac{X}{Y}\right\rceil is equal to \displaystyle\left\lfloor\frac{X+Y-1}{Y}\right\rfloor for 1 \leq X, Y
TIME COMPLEXITY
The time complexity is O(1) per test case.
SOLUTIONS
Setter's Solution
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
int main(){
ios_base::sync_with_stdio(false);
cin.tie();cout.tie();
long long TT = 1;
cin>>TT;
for(long long TR = 1;TR <= TT;TR++){
int n;
cin>>n;
int ans = (n+1)/2;
cout<<ans<<'\n';
}
return 0;
}
Tester's Solution 1
#include <bits/stdc++.h>
using namespace std;
/*
------------------------Input Checker----------------------------------
*/
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){
if(is_neg){
x= -x;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
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){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
/*
------------------------Main code starts here----------------------------------
*/
const int MAX_T = 1000;
const int MAX_N = 1000000000;
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
int sum_len=0;
void solve()
{
int n = readIntLn(1, MAX_N);
cout << n - n/2 << "\n";
}
signed main()
{
//fast;
#ifndef ONLINE_JUDGE
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
#endif
int t = readIntLn(1, MAX_T);
for(int i=1;i<=t;i++)
{
solve();
}
assert(getchar() == -1);
}
Tester's Solution 2
#include <bits/stdc++.h>
using namespace std;
/*
------------------------Input Checker----------------------------------
*/
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){
if(is_neg){
x= -x;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
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){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
/*
------------------------Main code starts here----------------------------------
*/
const int MAX_T = 1e5;
const int MAX_N = 1e5;
const int MAX_SUM_LEN = 1e5;
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ff first
#define ss second
#define mp make_pair
#define ll long long
#define rep(i,n) for(int i=0;i<n;i++)
#define rev(i,n) for(int i=n;i>=0;i--)
#define rep_a(i,a,n) for(int i=a;i<n;i++)
int sum_n = 0, sum_m = 0;
int max_n = 0, max_m = 0;
int yess = 0;
int nos = 0;
int total_ops = 0;
ll mod = 998244353;
ll po(ll x, ll n){
ll ans=1;
while(n>0){ if(n&1) ans=(ans*x)%mod; x=(x*x)%mod; n/=2;}
return ans;
}
void solve()
{
int n = readIntLn(1, 1e9);
max_n = max(max_n,n);
cout<<(n+1)/2<<'\n';
}
signed main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r" , stdin);
freopen("output.txt", "w" , stdout);
#endif
fast;
int t = 1;
t = readIntLn(1,1000);
for(int i=1;i<=t;i++)
{
solve();
}
//assert(getchar() == -1);
assert(sum_n<=1e6);
cerr<<"SUCCESS\n";
cerr<<"Tests : " << t << '\n';
cerr<<"Sum of lengths : " << sum_n <<'\n';
cerr<<"Maximum length : " << max_n <<'\n';
// cerr<<"Total operations : " << total_ops << '\n';
//cerr<<"Answered yes : " << yess << '\n';
//cerr<<"Answered no : " << nos << '\n';
}
Editorialist's Solution
import java.util.*;
import java.io.*;
class CHFDBT{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
pn((nl()+1)/2);
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static boolean multipleTC = true;
FastReader in;PrintWriter out;
void run() throws Exception{
in = new FastReader();
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 CHFDBT().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());}
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception{
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}catch (IOException e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception{
String str = "";
try{
str = br.readLine();
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
}
Feel free to share your approach. Suggestions are welcomed as always. 