给定一个大的正数作为字符串，计算给定数字的所有可被8整除的旋转。
例子：
Input: 8
Output: 1
Input: 40
Output: 1
Rotation: 40 is divisible by 8
04 is not divisible by 8
Input : 13502
Output : 0
No rotation is divisible by 8
Input : 43262488612
Output : 4方法：对于大数，很难将每个数字进行旋转和除以8。因此，使用了“ 8除数”属性，该属性表示如果将数字的最后3位除以8，则数字可以除以8。我们实际上并不旋转数字并检查最后8位数字的可除性，而是计算连续的3位数字(以循环方式)，这些数字可以被8整除。
插图：
Consider a number 928160
Its rotations are 928160, 092816, 609281,
160928, 816092, 281609.
Now form consecutive sequence of 3-digits from
the original number 928160 as mentioned in the
approach.
3-digit: (9, 2, 8), (2, 8, 1), (8, 1, 6),
(1, 6, 0),(6, 0, 9), (0, 9, 2)
We can observe that the 3-digit number formed by
the these sets, i.e., 928, 281, 816, 160, 609, 092,
are present in the last 3 digits of some rotation.
Thus, checking divisibility of these 3-digit numbers
gives the required number of rotations. C++
// C++ program to count all rotations divisible
// by 8
#include
using namespace std;
// function to count of all rotations divisible
// by 8
int countRotationsDivBy8(string n)
{
int len = n.length();
int count = 0;
// For single digit number
if (len == 1) {
int oneDigit = n[0] - '0';
if (oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers (considering all
// pairs)
if (len == 2) {
// first pair
int first = (n[0] - '0') * 10 + (n[1] - '0');
// second pair
int second = (n[1] - '0') * 10 + (n[0] - '0');
if (first % 8 == 0)
count++;
if (second % 8 == 0)
count++;
return count;
}
// considering all three-digit sequences
int threeDigit;
for (int i = 0; i < (len - 2); i++) {
threeDigit = (n[i] - '0') * 100 +
(n[i + 1] - '0') * 10 +
(n[i + 2] - '0');
if (threeDigit % 8 == 0)
count++;
}
// Considering the number formed by the
// last digit and the first two digits
threeDigit = (n[len - 1] - '0') * 100 +
(n[0] - '0') * 10 +
(n[1] - '0');
if (threeDigit % 8 == 0)
count++;
// Considering the number formed by the last
// two digits and the first digit
threeDigit = (n[len - 2] - '0') * 100 +
(n[len - 1] - '0') * 10 +
(n[0] - '0');
if (threeDigit % 8 == 0)
count++;
// required count of rotations
return count;
}
// Driver program to test above
int main()
{
string n = "43262488612";
cout << "Rotations: "
<< countRotationsDivBy8(n);
return 0;
} Java
// Java program to count all
// rotations divisible by 8
import java.io.*;
class GFG
{
// function to count of all
// rotations divisible by 8
static int countRotationsDivBy8(String n)
{
int len = n.length();
int count = 0;
// For single digit number
if (len == 1) {
int oneDigit = n.charAt(0) - '0';
if (oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers
// (considering all pairs)
if (len == 2) {
// first pair
int first = (n.charAt(0) - '0') *
10 + (n.charAt(1) - '0');
// second pair
int second = (n.charAt(1) - '0') *
10 + (n.charAt(0) - '0');
if (first % 8 == 0)
count++;
if (second % 8 == 0)
count++;
return count;
}
// considering all three-digit sequences
int threeDigit;
for (int i = 0; i < (len - 2); i++)
{
threeDigit = (n.charAt(i) - '0') * 100 +
(n.charAt(i + 1) - '0') * 10 +
(n.charAt(i + 2) - '0');
if (threeDigit % 8 == 0)
count++;
}
// Considering the number formed by the
// last digit and the first two digits
threeDigit = (n.charAt(len - 1) - '0') * 100 +
(n.charAt(0) - '0') * 10 +
(n.charAt(1) - '0');
if (threeDigit % 8 == 0)
count++;
// Considering the number formed by the last
// two digits and the first digit
threeDigit = (n.charAt(len - 2) - '0') * 100 +
(n.charAt(len - 1) - '0') * 10 +
(n.charAt(0) - '0');
if (threeDigit % 8 == 0)
count++;
// required count of rotations
return count;
}
// Driver program
public static void main (String[] args)
{
String n = "43262488612";
System.out.println( "Rotations: "
+countRotationsDivBy8(n));
}
}
// This code is contributed by vt_m.Python3
# Python3 program to count all
# rotations divisible by 8
# function to count of all
# rotations divisible by 8
def countRotationsDivBy8(n):
l = len(n)
count = 0
# For single digit number
if (l == 1):
oneDigit = int(n[0])
if (oneDigit % 8 == 0):
return 1
return 0
# For two-digit numbers
# (considering all pairs)
if (l == 2):
# first pair
first = int(n[0]) * 10 + int(n[1])
# second pair
second = int(n[1]) * 10 + int(n[0])
if (first % 8 == 0):
count+=1
if (second % 8 == 0):
count+=1
return count
# considering all
# three-digit sequences
threeDigit=0
for i in range(0,(l - 2)):
threeDigit = (int(n[i]) * 100 +
int(n[i + 1]) * 10 +
int(n[i + 2]))
if (threeDigit % 8 == 0):
count+=1
# Considering the number
# formed by the last digit
# and the first two digits
threeDigit = (int(n[l - 1]) * 100 +
int(n[0]) * 10 +
int(n[1]))
if (threeDigit % 8 == 0):
count+=1
# Considering the number
# formed by the last two
# digits and the first digit
threeDigit = (int(n[l - 2]) * 100 +
int(n[l - 1]) * 10 +
int(n[0]))
if (threeDigit % 8 == 0):
count+=1
# required count
# of rotations
return count
# Driver Code
if __name__=='__main__':
n = "43262488612"
print("Rotations:",countRotationsDivBy8(n))
# This code is contributed by mits.C#
// C# program to count all
// rotations divisible by 8
using System;
class GFG {
// function to count of all
// rotations divisible by 8
static int countRotationsDivBy8(String n)
{
int len = n.Length;
int count = 0;
// For single digit number
if (len == 1)
{
int oneDigit = n[0] - '0';
if (oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers
// (considering all pairs)
if (len == 2)
{
// first pair
int first = (n[0] - '0') *
10 + (n[1] - '0');
// second pair
int second = (n[1] - '0') *
10 + (n[0] - '0');
if (first % 8 == 0)
count++;
if (second % 8 == 0)
count++;
return count;
}
// considering all three -
// digit sequences
int threeDigit;
for (int i = 0; i < (len - 2); i++)
{
threeDigit = (n[i] - '0') * 100 +
(n[i + 1] - '0') * 10 +
(n[i + 2] - '0');
if (threeDigit % 8 == 0)
count++;
}
// Considering the number formed by the
// last digit and the first two digits
threeDigit = (n[len - 1] - '0') * 100 +
(n[0] - '0') * 10 +
(n[1] - '0');
if (threeDigit % 8 == 0)
count++;
// Considering the number formed
// by the last two digits and
// the first digit
threeDigit = (n[len - 2] - '0') * 100 +
(n[len - 1] - '0') * 10 +
(n[0] - '0');
if (threeDigit % 8 == 0)
count++;
// required count of rotations
return count;
}
// Driver Code
public static void Main ()
{
String n = "43262488612";
Console.Write("Rotations: "
+countRotationsDivBy8(n));
}
}
// This code is contributed by Nitin Mittal.PHP
Javascript
输出：
Rotations: 4时间复杂度： O(n)，其中n是输入数字中的位数。