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