给定一个大的正数作为字符串，计算给定数的所有可被4整除的旋转。
例子：
Input: 8
Output: 1
Input: 20
Output: 1
Rotation: 20 is divisible by 4
02 is not divisible by 4
Input : 13502
Output : 0
No rotation is divisible by 4
Input : 43292816
Output : 5
5 rotations are : 43292816, 16432928, 81643292
92816432, 32928164 对于大数，很难将每个数字进行旋转和除以4。因此，使用了“ 4除数”属性，该属性表示如果将数字的最后2位除以4，则数字可以被4除。实际不旋转数字并检查最后2位数字的可除性，而是我们将连续对(以循环方式)计数为可被4整除的对。
插图：
Consider a number 928160
Its rotations are 928160, 092816, 609281, 160928,
816092, 281609.
Now form pairs from the original number 928160
as mentioned in the approach.
Pairs: (9,2), (2,8), (8,1), (1,6),
(6,0), (0,9)
We can observe that the 2-digit number formed by the these
pairs, i.e., 92, 28, 81, 16, 60, 09, are present in the last
2 digits of some rotation.
Thus, checking divisibility of these pairs gives the required
number of rotations.
Note: A single digit number can directly
be checked for divisibility.下面是该方法的实现。
C++
// C++ program to count all rotation divisible
// by 4.
#include
using namespace std;
// Returns count of all rotations divisible
// by 4
int countRotations(string n)
{
int len = n.length();
// For single digit number
if (len == 1)
{
int oneDigit = n.at(0)-'0';
if (oneDigit%4 == 0)
return 1;
return 0;
}
// At-least 2 digit number (considering all
// pairs)
int twoDigit, count = 0;
for (int i=0; i<(len-1); i++)
{
twoDigit = (n.at(i)-'0')*10 + (n.at(i+1)-'0');
if (twoDigit%4 == 0)
count++;
}
// Considering the number formed by the pair of
// last digit and 1st digit
twoDigit = (n.at(len-1)-'0')*10 + (n.at(0)-'0');
if (twoDigit%4 == 0)
count++;
return count;
}
//Driver program
int main()
{
string n = "4834";
cout << "Rotations: " << countRotations(n) << endl;
return 0;
} Java
// Java program to count
// all rotation divisible
// by 4.
import java.io.*;
class GFG {
// Returns count of all
// rotations divisible
// by 4
static int countRotations(String n)
{
int len = n.length();
// For single digit number
if (len == 1)
{
int oneDigit = n.charAt(0)-'0';
if (oneDigit % 4 == 0)
return 1;
return 0;
}
// At-least 2 digit
// number (considering all
// pairs)
int twoDigit, count = 0;
for (int i = 0; i < (len-1); i++)
{
twoDigit = (n.charAt(i)-'0') * 10 +
(n.charAt(i+1)-'0');
if (twoDigit%4 == 0)
count++;
}
// Considering the number
// formed by the pair of
// last digit and 1st digit
twoDigit = (n.charAt(len-1)-'0') * 10 +
(n.charAt(0)-'0');
if (twoDigit%4 == 0)
count++;
return count;
}
//Driver program
public static void main(String args[])
{
String n = "4834";
System.out.println("Rotations: " +
countRotations(n));
}
}
// This code is contributed by Nikita tiwari.Python3
# Python3 program to count
# all rotation divisible
# by 4.
# Returns count of all
# rotations divisible
# by 4
def countRotations(n) :
l = len(n)
# For single digit number
if (l == 1) :
oneDigit = (int)(n[0])
if (oneDigit % 4 == 0) :
return 1
return 0
# At-least 2 digit number
# (considering all pairs)
count = 0
for i in range(0, l - 1) :
twoDigit = (int)(n[i]) * 10 + (int)(n[i + 1])
if (twoDigit % 4 == 0) :
count = count + 1
# Considering the number
# formed by the pair of
# last digit and 1st digit
twoDigit = (int)(n[l - 1]) * 10 + (int)(n[0])
if (twoDigit % 4 == 0) :
count = count + 1
return count
# Driver program
n = "4834"
print("Rotations: " ,
countRotations(n))
# This code is contributed by Nikita tiwari.C#
// C# program to count all rotation
// divisible by 4.
using System;
class GFG {
// Returns count of all
// rotations divisible
// by 4
static int countRotations(String n)
{
int len = n.Length;
// For single digit number
if (len == 1)
{
int oneDigit = n[0] - '0';
if (oneDigit % 4 == 0)
return 1;
return 0;
}
// At-least 2 digit
// number (considering all
// pairs)
int twoDigit, count = 0;
for (int i = 0; i < (len - 1); i++)
{
twoDigit = (n[i] - '0') * 10 +
(n[i + 1] - '0');
if (twoDigit % 4 == 0)
count++;
}
// Considering the number
// formed by the pair of
// last digit and 1st digit
twoDigit = (n[len - 1] - '0') * 10 +
(n[0] - '0');
if (twoDigit % 4 == 0)
count++;
return count;
}
//Driver program
public static void Main()
{
String n = "4834";
Console.Write("Rotations: " +
countRotations(n));
}
}
// This code is contributed by nitin mittal.PHP
Javascript
输出：
Rotations: 2时间复杂度： O(n)，其中n是输入数字中的位数。