给定一个整数N ，任务是在除去一些数字(可能没有数字)之后，将数字减小为最小的正整数X ，以使X可以被4整除。如果不能将其减小到-1，则打印-1 。
例子：
Input: N = 78945666384
Output: 4
Remove all the digits except a single
occurrence of the digit ‘4’.
Input: N = 17
Output: -1
方法：由于必须将结果数量最小化。因此，请检查数字中是否有任何等于“ 4”或“ 8”的数字，因为这些数字是按升序可被4整除的数字。如果没有这样的数字，则检查长度为2的数字的所有子序列是否为4的倍数。如果仍然没有4的倍数，则该数字是不可能的，因为任何2位数以上的数字都是4的倍数，那么肯定会有一个被4整除的子序列，而数字要少3 。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
const int TEN = 10;
// Function to return the minimum number
// that can be formed after removing
// the digits which is a multiple of 4
int minNum(string str, int len)
{
int res = INT_MAX;
// For every digit of the number
for (int i = 0; i < len; i++) {
// Check if the current digit
// is divisible by 4
if (str[i] == '4' || str[i] == '8') {
res = min(res, str[i] - '0');
}
}
for (int i = 0; i < len - 1; i++) {
for (int j = i + 1; j < len; j++) {
int num = (str[i] - '0') * TEN
+ (str[j] - '0');
// If any subsequence of two
// digits is divisible by 4
if (num % 4 == 0) {
res = min(res, num);
}
}
}
return ((res == INT_MAX) ? -1 : res);
}
// Driver code
int main()
{
string str = "17";
int len = str.length();
cout << minNum(str, len);
return 0;
} Java
// Java implementation of the approach
class GFG
{
static int TEN = 10;
// Function to return the minimum number
// that can be formed after removing
// the digits which is a multiple of 4
static int minNum(char[] str, int len)
{
int res = Integer.MAX_VALUE;
// For every digit of the number
for (int i = 0; i < len; i++)
{
// Check if the current digit
// is divisible by 4
if (str[i] == '4' || str[i] == '8')
{
res = Math.min(res, str[i] - '0');
}
}
for (int i = 0; i < len - 1; i++)
{
for (int j = i + 1; j < len; j++)
{
int num = (str[i] - '0') * TEN
+ (str[j] - '0');
// If any subsequence of two
// digits is divisible by 4
if (num % 4 == 0)
{
res = Math.min(res, num);
}
}
}
return ((res == Integer.MAX_VALUE) ? -1 : res);
}
// Driver code
public static void main(String[] args)
{
String str = "17";
int len = str.length();
System.out.print(minNum(str.toCharArray(), len));
}
}
// This code is contributed by 29AjayKumarPython 3
# Python3 implementation of the approach
import sys
TEN = 10
# Function to return the minimum number
# that can be formed after removing
# the digits which is a multiple of 4
def minNum(str, len1):
res = sys.maxsize
# For every digit of the number
for i in range(len1):
# Check if the current digit
# is divisible by 4
if (str[i] == '4' or str[i] == '8'):
res = min(res, ord(str[i]) - ord('0'))
for i in range(len1 - 1):
for j in range(i + 1, len1, 1):
num = (ord(str[i]) - ord('0')) * TEN + \
(ord(str[j]) - ord('0'))
# If any subsequence of two
# digits is divisible by 4
if (num % 4 == 0):
res = min(res, num)
if (res == sys.maxsize):
return -1
else:
return res
# Driver code
if __name__ == '__main__':
str = "17"
len1 = len(str)
print(minNum(str, len1))
# This code is contributed by Surendra_GangwarC#
// C# implementation of the approach
using System;
class GFG
{
static int TEN = 10;
// Function to return the minimum number
// that can be formed after removing
// the digits which is a multiple of 4
static int minNum(char[] str, int len)
{
int res = int.MaxValue;
// For every digit of the number
for (int i = 0; i < len; i++)
{
// Check if the current digit
// is divisible by 4
if (str[i] == '4' || str[i] == '8')
{
res = Math.Min(res, str[i] - '0');
}
}
for (int i = 0; i < len - 1; i++)
{
for (int j = i + 1; j < len; j++)
{
int num = (str[i] - '0') * TEN
+ (str[j] - '0');
// If any subsequence of two
// digits is divisible by 4
if (num % 4 == 0)
{
res = Math.Min(res, num);
}
}
}
return ((res == int.MaxValue) ? -1 : res);
}
// Driver code
public static void Main(String[] args)
{
String str = "17";
int len = str.Length;
Console.Write(minNum(str.ToCharArray(), len));
}
}
// This code is contributed by Rajput-Ji输出：
-1