**Counter game**

### Problem Statement :

Louise and Richard have developed a numbers game. They pick a number and check to see if it is a power of 2. If it is, they divide it by 2. If not, they reduce it by the next lower number which is a power of 2. Whoever reduces the number to 1 wins the game. Louise always starts. Given an initial value, determine who wins the game. Example n = 132 It's Louise's turn first. She determines that 132 is not a power of 2. The next lower power of 2 is 128, so she subtracts that from 132 and passes 4 to Richard. 4 is a power of 2, so Richard divides it by 2 and passes 2 to Louise. Likewise, 2 is a power so she divides it by 2 and reaches 1. She wins the game. Update If they initially set counter to 1, Richard wins. Louise cannot make a move so she loses. Function Description Complete the counterGame function in the editor below. counterGame has the following parameter(s): int n: the initial game counter value Returns string: either Richard or Louise Input Format The first line contains an integer t, the number of testcases. Each of the next t lines contains an integer n, the initial value for each game. Constraints 1 <= t <= 10 1 <= n <= 2^64-1

### Solution :

` ````
Solution in C :
In C++ :
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
int t;
unsigned long long int n;
cin>>t;
while(t--)
{
int c=0;
cin>>n;
//if(n==1) {cout<<"Louise";continue;}
while(n>1)
{
if((n&(n-1))==0)
{ n/=2;
c++;
}
else
{
unsigned long long int i=n;
int size=0;
while(i>0)
{ i=i>>1;size++;
}
n=n-(1ULL<<(size-1));
c++;
}
}
if(c%2!=0) cout<<"Louise\n";
else cout<<"Richard\n";
}
return 0;
}
In Java :
import java.io.*;
import java.math.BigInteger;
import java.util.*;
public class Solution {
static BufferedReader reader;
static StringTokenizer tokenizer = null;
static PrintWriter writer;
static String nextToken() throws IOException {
while (tokenizer == null || (!tokenizer.hasMoreTokens())) {
tokenizer = new StringTokenizer(reader.readLine());
}
return tokenizer.nextToken();
}
static int nextInt() throws NumberFormatException, IOException {
return Integer.parseInt(nextToken());
}
static double nextDouble() throws NumberFormatException, IOException {
return Double.parseDouble(nextToken());
}
static long nextLong() throws NumberFormatException, IOException {
return Long.parseLong(nextToken());
}
public static void main(String[] args) throws IOException {
reader = new BufferedReader(new InputStreamReader(System.in));
writer = new PrintWriter(System.out);
cherry();
reader.close();
writer.close();
}
static BigInteger two = new BigInteger("2");
static boolean isWin(BigInteger bi) {
if (bi.equals(BigInteger.ONE)) {
return false;
}
if (bi.bitCount() == 1) {
return !isWin(bi.divide(two));
} else {
return !isWin(bi.clearBit(bi.bitLength() - 1));
}
}
static void cherry() throws NumberFormatException, IOException {
int T = nextInt();
for (int t = 0; t < T; t++) {
BigInteger bi = new BigInteger(nextToken());
writer.println(isWin(bi) ? "Louise" : "Richard");
}
}
}
In C :
#include<stdio.h>
#include<math.h>
int main(){
int nt;
scanf("%d",&nt);
int x,y;
long long unsigned int n;
for(x=0;x<nt;x++){
scanf("%llu",&n);
int bitcount = 0;
long long unsigned int value=n;
while (value != 0){
bitcount++;
value &= value - 1;
}
bitcount=bitcount-1;
int loop_count=0;
long long unsigned int temp=1;
while(!(n&temp)){
loop_count++;
temp=pow(2,loop_count);
}
if((bitcount+loop_count)%2==0){
printf("Richard\n");
}
else{
printf("Louise\n");
}
}
return 0;
}
In Python3 :
import math
from math import log2
from math import floor
def genlist(x):
if x==1:
return False
elif (log2(x))%1==0:
return not genlist(int(x/2))
else:
return not genlist(x-2**floor(log2(x)))
tcs=int(input())
for tc in range(tcs):
sa=int(input())
if genlist(sa):
print("Louise")
else:
print("Richard")
```

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