MEX-OR MEXOR Codechef October Long Challenge

MEX-OR MEXOR Solution

The MEX (minimum excluded) of an array is the smallest non-negative integer that does not belong to the array. For instance:

  • The MEX of [2,2,1] is 0, because 0 does not belong to the array.
  • The MEX of [3,1,0,1] is 2, because 0 and 1 belong to the array, but 2 does not.
  • The MEX of [0,3,1,2] is 4 because 0,1,2 and 3 belong to the array, but 4 does not.

Find the maximum possible MEX of an array of non-negative integers such that the bitwise OR of the elements in the array does not exceed X.

Input Format

  • The first line contains T denoting the number of test cases. Then the test cases follow.
  • Each test case contains a single integer X on a single line.

Output Format

For each test case, output on a single line the maximum possible MEX of the array satisfying the given property.

Constraints

  • 1≤T≤105
  • 0≤X≤109

Subtasks

Subtask 1 (100 points): Original constraints

Sample Input 1

4
0
1
2
5

Sample Output 1

1
2
2
4

Explanation

Test case 1: The array could be [0].

Test case 2: The array could be [0,1]. Here the bitwise OR of 0 and 1 is 1 and the MEX of the array is 2 as both 0 and 1 belongs to the array.

Test case 4: The array could be [1,0,3,2]. Here the bitwise OR of all integers in the array is 3 and the MEX of the array is 4.

SOLUTION

Program C++: MEX-OR MEXOR Solution in C++

#include <iostream>
#include<cmath>
using namespace std;
int main() {
  // your code goes here
  int t; cin>>t;
  while(t--){
      int m;
      cin>>m;
      bool flag=true;
      int count=0,x=m;
      if(x==0){
      cout<<"1"<<endl;
      continue;
      }
      while(x!=0)
      {
          int l=x%2;
          if(l==0)
          flag=false;
          
          x/=2;
          count++;
      }
     
      if(flag) 
      cout<<m+1<<endl;
       else{
          int sum=0;
          count--;
          while(count--){
              sum+=pow(2,count);
          }
          cout<<sum+1<<endl;
      }
  }
  return 0;
}

October Long Challenge 2021

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