在C程序中，将一个数组中具有最大AND值的一对元素打印出来

根据问题，我们给定了一个包含n个正整数的数组，我们需要从数组中找到具有最大AND值的一对。

示例

Input: arr[] = { 4, 8, 12, 16 }
Output: pair = 8 12
The maximum and value= 8

Input:arr[] = { 4, 8, 16, 2 }
Output: pair = No possible AND
The maximum and value = 0

寻找最大AND值的方法类似于在数组中寻找最大AND值。程序必须找到导致得到的AND值的元素对。为了找到这些元素，只需遍历整个数组，并找到每个元素与得到的最大AND值（结果）的AND值，如果arr[i] & result == result，则表示arr[i]是将生成最大AND值的元素。此外，在最大AND值（结果）为零的情况下，我们应该打印“不可能”。

算法

int checkBit(int pattern, int arr[], int n)
START
STEP 1: DECLARE AND INITIALIZE count AS 0
STEP 2: LOOP FOR i = 0 AND i < n AND i++
   IF (pattern & arr[i]) == pattern THEN,
      INCREMENT count BY 1
STEP 3: RETURN count
STOP
int maxAND(int arr[], int n)
START
STEP 1: DECLARE AND INITIALIZE res = 0 AND count
STEP 2: LOOP FOR bit = 31 AND bit >= 0 AND bit--
   count = GOTO FUNCTION checkBit(res | (1 << bit), arr,n)
   IF count >= 2 THEN,
      res |= (1 << bit);
   END IF
   IF res == 0
      PRINT "no possible AND”
   ELSE
      PRINT "Pair with maximum AND= "
   count = 0;
   LOOP FOR int i = 0 AND i < n && count < 2 AND i++
      IF (arr[i] & res) == res THEN,
         INCREMENT count BY 1
         PRINT arr[i]
      END IF
   END FOR
END FOR
RETURN res
STOP

Example

示例

#include 
int checkBit(int pattern, int arr[], int n){
   int count = 0;
   for (int i = 0; i < n; i++)
      if ((pattern & arr[i]) == pattern)
         count++;
   return count;
}
// Function for finding maximum AND value pair
int maxAND(int arr[], int n){
   int res = 0, count;
   for (int bit = 31; bit >= 0; bit--) {
      count = checkBit(res | (1 << bit), arr, n);  
      if (count >= 2)
         res |= (1 << bit);
   }
   if (res == 0) //if there is no pair available
      printf("no possible and
");
   else { //Printing the pair available
      printf("Pair with maximum AND= ");
      count = 0;
      for (int i = 0; i < n && count < 2; i++) {
         // incremnent count value after
         // printing element
         if ((arr[i] & res) == res) {
            count++;
            printf("%d ", arr[i]);
         }
      }
   }
   return res;
}
int main(int argc, char const *argv[]){
   int arr[] = {5, 6, 2, 8, 9, 12};
   int n = sizeof(arr)/sizeof(arr[0]);
   int ma = maxAND(arr, n);
   printf("
The maximum AND value= %d ", ma);
   return 0;
}

输出

如果我们运行上述程序，它将生成以下输出−

pair = 8 9
The maximum and value= 8