在C程序的数组中以最大AND值打印对。
根据这个问题，我们得到一个n个正整数的数组，我们必须从该数组中找到一个具有最大AND值的对。
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查找最大AND值的方法类似于在数组中查找最大AND值。程序必须找出导致获得AND值的一对元素。为了找到元素，只需遍历整个数组并找到具有最大AND值（结果）的每个元素的AND值，如果arr[i]＆result==result，则意味着arr[i]是将产生最大AND值。同样，如果最大AND值（结果）为零，则在这种情况下，我们应打印“Not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示例
#include <stdio.h>
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;
}
//查找最大AND值对的功能
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\n");
else { //Printing the pair available
printf("Pair with maximum AND= ");
count = 0;
for (int i = 0; i < n && count < 2; i++) {
//之后的递增计数值
//打印元素
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("\nThe maximum AND value= %d ", ma);
return 0;
}输出结果
如果我们运行上面的程序，那么它将生成以下输出-
pair = 8 9 The maximum and value= 8