给定八进制数N。任务是编写一个程序来检查给定八进制数N的十进制表示形式是否可被7整除。
例子:
Input: N = 112
Output: NO
Equivalent Decimal = 74
7410 = 7 * 10 1 + 4 * 100
1128 = 1 * 82 + 1 * 81 + 2 * 80
Input: N = 25
Output: YES
Decimal Equivalent = 21
这个想法是要注意的是,8%7将返回1。因此,当我们扩展八进制表示形式并将其取模7时,单个术语8的所有幂将减少为1。因此,如果八进制表示形式中所有数字的总和可以被7整除,那么相应的十进制数将被7整除。
下面是上述方法的实现:
C++
// CPP program to check if Decimal representation
// of an Octal number is divisible by 7 or not
#include
using namespace std;
// Function to check Divisibility
int check(int n)
{
int sum = 0;
// Sum of all individual digits
while (n != 0) {
sum += n % 10;
n = n / 10;
}
// Condition
if (sum % 7 == 0)
return 1;
else
return 0;
}
// Driver Code
int main()
{
// Octal number
int n = 25;
(check(n) == 1) ? cout << "YES"
: cout << "NO";
return 0;
}
Java
// Java program to check if Decimal
// representation of an Octal number
// is divisible by 7 or not
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
// Function to check Divisibility
static int check(int n)
{
int sum = 0;
// Sum of all individual digits
while (n != 0)
{
sum += n % 10;
n = n / 10;
}
// Condition
if (sum % 7 == 0)
return 1;
else
return 0;
}
// Driver Code
public static void main(String args[])
{
// Octal number
int n = 25;
String s=(check(n) == 1) ?
"YES" : "NO";
System.out.println(s);
}
}
// This code is contributed
// by Subhadeep
Python 3
# Python 3 program to check if
# Decimal representation of an
# Octal number is divisible by
# 7 or not
# Function to check Divisibility
def check(n):
sum = 0
# Sum of all individual digits
while n != 0 :
sum += n % 10
n = n // 10
# Condition
if sum % 7 == 0 :
return 1
else:
return 0
# Driver Code
if __name__ == "__main__":
# Octal number
n = 25
print(("YES") if check(n) == 1
else print("NO"))
# This code is contributed
# by ChitraNayal
C#
// C# program to check if Decimal
// representation of an Octal
// number is divisible by 7 or not
using System;
class GFG
{
// Function to check Divisibility
static int check(int n)
{
int sum = 0;
// Sum of all individual digits
while (n != 0)
{
sum += n % 10;
n = n / 10;
}
// Condition
if (sum % 7 == 0)
return 1;
else
return 0;
}
// Driver Code
public static void Main(String []args)
{
// Octal number
int n = 25;
String s=(check(n) == 1) ?
"YES" : "NO";
Console.WriteLine(s);
}
}
// This code is contributed
// by Kirti_Mangal
PHP
Javascript
输出:
YES