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  1. Prepare
  2. C
  3. Conditionals and Loops
  4. Bitwise Operators
  5. Discussions

Bitwise Operators

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569 Discussions

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  • mohammedasifsha2
     + 0 comments

    { int max_and = 0, max_or = 0, max_xor = 0;

    for (int a = 1; a <= n; a++) {
    for (int b = a + 1; b <= n; b++) {
        int and_val = a & b;
        int or_val = a | b;
        int xor_val = a ^ b;

        if (and_val > max_and && and_val < k) {
            max_and = and_val;
        }
        if (or_val > max_or && or_val < k) {
            max_or = or_val;
        }
        if (xor_val > max_xor && xor_val < k) {
            max_xor = xor_val;
        }
    }
}

printf("%d\n%d\n%d\n", max_and, max_or, max_xor);

    }

  • u2404126
     + 0 comments

    include

    int main() { int n,k; scanf("%d %d", &n, &k); int b=2, a=1,real_b = 0, max_or=0, max_and=0,max_xor=0; for(a; a<=n; a++){ if(a==b){ b++; } real_b=b; for(b; b<=n; b++){ int and=0, or=0, xor=0; and = a&b; or = a | b; xor = a ^ b; if(and>max_and && andmax_or && ormax_xor && xor < k){ max_xor = xor; }

        }
    b = real_b;
}
printf("%d\n%d\n%d", max_and,max_or,max_xor);
return 0;

    }

  • inyeobgim
     + 0 comments
    void calculate_the_maximum(int n, int k) {
    
  //Write your code here.
  int numbersToConsider[n];
	//Numbers in answer array will represent biggest and, or, xor in sequence
  int answer[3] = {0, 0, 0};
  
  //Making an array of numbers from 1 to n
  for(int i = 0; i < n; i++){
    numbersToConsider[i] = i+1;
  } 
  
  for(int i = 0; i < numbersToConsider[n-1]; i++){
    for(int j = 1; numbersToConsider[i] + j <= n ; j++){
        int and;
        and = numbersToConsider[i] & numbersToConsider[i+j];
        if(and > answer[0] && and < k){
            answer[0] = and;
        }
        int or;
        or = numbersToConsider[i] | numbersToConsider[i+j];
        if(or > answer[1] && or < k){
            answer[1] = or;
        }
        int xor;
        xor = numbersToConsider[i] ^ numbersToConsider[i+j];
        if(xor > answer[2] && xor < k){
            answer[2] = xor;
        }
    }
  }
  printf("%d\n%d\n%d", answer[0], answer[1], answer[2]);
}

  • srijakuruba2005
     + 0 comments

    void calculate_the_maximum(int n, int k) { int i,j,and,or,xor,M1=0,M2=0,M3=0; for(i=1;i 

    }

    } printf("%d\n",M1); printf("%d\n",M2); printf("%d\n",M3); }

  • akulbachandra67
     + 0 comments

    **Mine C solution **

    void calculate_the_maximum(int n, int k) {
  int a=0,b=0,c=0; 
  for(int i=1; i<= n; i++){
    for(int j= i+1; j<=n; j++){
        if (((i&j)< k) && ((i&j)> a)){ a = (i&j); } 
        if (((i|j)< k) && ((i|j)> b)){ b = (i|j); } 
        if (((i^j)< k) && ((i^j)> c)){ c = (i^j); } 
    }
  }
  printf("%d\n%d\n%d",a,b,c);
}