检查数字是否能被 43 整除
原文:https://www . geesforgeks . org/check-如果数字可被 43 整除或不可被整除/
给定一个数 N ，任务是检查这个数是否能被 43 整除。 例:
输入: N = 2795 输出:是 解释: 43 * 65 = 2795 输入: N = 11094 输出:是 解释: 43 * 258 = 11094
方法:43 的可除性检验为:
- 提取最后一位数字。
- 从去掉最后一位数字后得到的剩余数字中加上 13 *最后一位数字。
- 重复上述步骤，直到得到一个两位数或零。
- 如果两位数可以被 43 整除，或者是 0，那么原数也可以被 43 整除。
例如:
If N = 11739
Step 1:
N = 11739
Last digit = 9
Remaining number = 1173
Adding 13 times last digit
Resultant number = 1173 + 13*9 = 1290
Step 2:
N = 1290
Since 129 is divisible by 43 as 43 * 3 = 129
Therefore N = 11739 is also divisible by 43
以下是上述方法的实现:
C++
// C++ program to check whether a number
// is divisible by 43 or not
#include<bits/stdc++.h>
#include<stdlib.h>
using namespace std;
// Function to check if the number is divisible by 43 or not
bool isDivisible(int n)
{
int d;
// While there are at least two digits
while (n / 100)
{
// Extracting the last
d = n % 10;
// Truncating the number
n /= 10;
// adding thirteen times the last
// digit to the remaining number
n = abs(n+(d * 13));
}
// Finally return if the two-digit
// number is divisible by 43 or not
return (n % 43 == 0) ;
}
// Driver Code
int main() {
int N = 2795;
if (isDivisible(N))
cout<<"Yes"<<endl ;
else
cout<<"No"<<endl ;
return 0;
}
// This code is contributed by ANKITKUMAR34
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to check whether a number
// is divisible by 43 or not
class GFG
{
// Function to check if the number is divisible by 43 or not
static boolean isDivisible(int n)
{
int d;
// While there are at least two digits
while ((n / 100) > 0)
{
// Extracting the last
d = n % 10;
// Truncating the number
n /= 10;
// adding thirteen times the last
// digit to the remaining number
n = Math.abs(n+(d * 13));
}
// Finally return if the two-digit
// number is divisible by 43 or not
return (n % 43 == 0) ;
}
// Driver Code
public static void main(String[] args) {
int N = 2795;
if (isDivisible(N))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by PrinciRaj1992
Python 3
# Python program to check whether a number
# is divisible by 43 or not
# Function to check if the number is
# divisible by 43 or not
def isDivisible(n) :
# While there are at least two digits
while n // 100 :
# Extracting the last
d = n % 10
# Truncating the number
n //= 10
# Adding thirteen times the last
# digit to the remaining number
n = abs(n+(d * 13))
# Finally return if the two-digit
# number is divisible by 43 or not
return (n % 43 == 0)
# Driver Code
if __name__ == "__main__" :
N = 2795
if (isDivisible(N)):
print("Yes")
else :
print("No")
C
// C# program to check whether a number
// is divisible by 43 or not
using System;
class GFG
{
// Function to check if the number is divisible by 43 or not
static bool isDivisible(int n)
{
int d;
// While there are at least two digits
while (n / 100 > 0)
{
// Extracting the last
d = n % 10;
// Truncating the number
n /= 10;
// adding thirteen times the last
// digit to the remaining number
n = Math.Abs(n + (d * 13));
}
// Finally return if the two-digit
// number is divisible by 43 or not
return (n % 43 == 0) ;
}
// Driver Code
public static void Main()
{
int N = 2795;
if (isDivisible(N))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by AbhiThakur
java 描述语言
<script>
//javascript program to check whether a number
// is divisible by 43 or not
// Function to check if the number is divisible by 43 or not
function isDivisible(n)
{
let d;
// While there are at least two digits
while(parseInt(n/100) > 0)
{
// Extracting the last
d = n % 10;
// Truncating the number
n = parseInt(n / 10)
// adding thirteen times the last
// digit to the remaining number
n = Math.abs(n+(d * 13));
}
// Finally return if the two-digit
// number is divisible by 43 or not
return (n % 43 == 0) ;
}
// Driver Code
let N = 2795;
if (isDivisible(N))
document.write("Yes");
else
document.write("No");
// This code is contributed by vaibhavrabadiya117.
</script>
Output:
Yes
时间复杂度:O(log 10 N )
辅助空间:0(1)
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