数组的每个子集的元素的“与”之间的最小值
给定一个整数数组，任务是找到数组每个子集的所有元素的“与”，并打印所有这些元素中的最小“与”值。 例:
Input: arr[] = {1, 2, 3}
Output: 0
AND of all possible subsets
(1 & 2) = 0,
(1 & 3) = 1,
(2 & 3) = 2 and
(1 & 2 & 3) = 0\.
Minimum among these is 0\.
Input: arr[] = {7, 2}
Output: 2
方法:数组任何子集的最小 AND 值将是数组所有元素的 AND。所以，最简单的方法就是求子数组所有元素的“与”。 以下是上述办法的实施: 实施:
C++
// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
void minAND(int arr[], int n)
{
int s = arr[0];
// Find AND of whole array
for (int i = 1; i < n; i++)
{
s = s & arr[i];
}
// Print the answer
cout << (s) << endl;
}
// Driver code
int main()
{
int arr[] = {1, 2, 3};
int n = sizeof(arr)/sizeof(int);
minAND(arr, n);
}
// This code has been contributed by Arnab Kundu
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
class GFG
{
static void minAND(int[] arr, int n)
{
int s = arr[0];
// Find AND of whole array
for (int i = 1; i < n; i++)
{
s = s & arr[i];
}
// Print the answer
System.out.println(s);
}
// Driver code
public static void main(String[] args)
{
int[] arr = {1, 2, 3};
int n = arr.length;
minAND(arr, n);
}
}
// This code has been contributed by 29AjayKumar
计算机编程语言
# Python program for the above approach
def minAND(arr, n):
s = arr[0]
# Find AND of whole array
for i in range(1, n):
s = s & arr[i]
# Print the answer
print(s)
# Driver code
arr = [1, 2, 3]
n = len(arr)
minAND(arr, n)
C
// C# program for the above approach
class GFG
{
static void minAND(int[] arr, int n)
{
int s = arr[0];
// Find AND of whole array
for (int i = 1; i < n; i++)
{
s = s & arr[i];
}
// Print the answer
System.Console.WriteLine(s);
}
// Driver code
static void Main()
{
int[] arr = {1, 2, 3};
int n = arr.Length;
minAND(arr, n);
}
}
// This code has been contributed by chandan_jnu
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program for the above approach
function minAND($arr, $n)
{
$s = $arr[0];
// Find AND of whole array
for ($i = 1; $i < $n; $i++)
{
$s = $s & $arr[$i];
}
// Print the answer
print($s . "\n");
}
// Driver code
$arr = array(1, 2, 3);
$n = count($arr);
minAND($arr, $n);
// This code is contributed
// by chandan_jnu
?>
java 描述语言
<script>
// Javascript implementation of the approach
function minAND(arr, n) {
let s = arr[0];
// Find AND of whole array
for (let i = 1; i < n; i++) {
s = s & arr[i];
}
// Prlet the answer
document.write((s) + "<br>");
}
// Driver code
let arr = [1, 2, 3];
let n = arr.length;
minAND(arr, n);
// This code is contributed by _saurabh_jaiswal
</script>
Output:
0
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