检查数字的位数总和是否能被其所有位数整除
给定一个整数 N ，任务是检查给定数的位数之和是否能被其所有位数整除。如果可分，则打印是否则打印否。
示例:
输入: N = 12 输出:否 位数之和= 1 + 2 = 3 3 可被 1 整除但不能被 2 整除。
输入:N = 123 T3】输出:是
方法:先求数字的位数之和，然后逐一检查，计算出的和是否能被数字的所有位数整除。如果某个数字不能被整除，则打印否否则打印是。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if all the digits
// of n divide the sum of the digits of n
bool isDivisible(long long int n)
{
// Store a copy of the original number
long long int temp = n;
// Find the sum of the digits of n
int sum = 0;
while (n) {
int digit = n % 10;
sum += digit;
n /= 10;
}
// Restore the original value
n = temp;
// Check if all the digits divide
// the calculated sum
while (n) {
int digit = n % 10;
// If current digit doesn't
// divide the sum
if (sum % digit != 0)
return false;
n /= 10;
}
return true;
}
// Driver code
int main()
{
long long int n = 123;
if (isDivisible(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function that returns true if all the digits
// of n divide the sum of the digits of n
static boolean isDivisible(long n)
{
// Store a copy of the original number
long temp = n;
// Find the sum of the digits of n
int sum = 0;
while (n != 0)
{
int digit = (int) n % 10;
sum += digit;
n /= 10;
}
// Restore the original value
n = temp;
// Check if all the digits divide
// the calculated sum
while (n != 0)
{
int digit = (int)n % 10;
// If current digit doesn't
// divide the sum
if (sum % digit != 0)
return false;
n /= 10;
}
return true;
}
// Driver code
public static void main (String[] args)
{
long n = 123;
if (isDivisible(n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by AnkitRai01
计算机编程语言
# Python implementation of the approach
# Function that returns true if all the digits
# of n divide the sum of the digits of n
def isDivisible(n):
# Store a copy of the original number
temp = n
# Find the sum of the digits of n
sum = 0
while (n):
digit = n % 10
sum += digit
n //= 10
# Restore the original value
n = temp
# Check if all the digits divide
# the calculated sum
while(n):
digit = n % 10
# If current digit doesn't
# divide the sum
if(sum % digit != 0):
return False
n //= 10;
return True
# Driver code
n = 123
if(isDivisible(n)):
print("Yes")
else:
print("No")
C
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true if all the digits
// of n divide the sum of the digits of n
static bool isDivisible(long n)
{
// Store a copy of the original number
long temp = n;
// Find the sum of the digits of n
int sum = 0;
while (n != 0)
{
int digit = (int) n % 10;
sum += digit;
n /= 10;
}
// Restore the original value
n = temp;
// Check if all the digits divide
// the calculated sum
while (n != 0)
{
int digit = (int)n % 10;
// If current digit doesn't
// divide the sum
if (sum % digit != 0)
return false;
n /= 10;
}
return true;
}
// Driver code
static public void Main ()
{
long n = 123;
if (isDivisible(n))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by @tushil.
java 描述语言
<script>
// Javascript implementation of the approach
// Function that returns true if all the
// digits of n divide the sum of the digits
// of n
function isDivisible(n)
{
// Store a copy of the original number
var temp = n;
// Find the sum of the digits of n
var sum = 0;
while (n != 0)
{
var digit = parseInt(n % 10);
sum += digit;
n = parseInt(n / 10);
}
// Restore the original value
n = temp;
// Check if all the digits divide
// the calculated sum
while (n != 0)
{
var digit = parseInt(n % 10);
// If current digit doesn't
// divide the sum
if (sum % digit != 0)
return false;
n = parseInt(n/10);
}
return true;
}
// Driver code
var n = 123;
if (isDivisible(n))
document.write("Yes");
else
document.write("No");
// This code is contributed by todaysgaurav
</script>
Output
Yes
时间复杂度: O(log(N)) 辅助空间: O(1)
方法 2:使用字符串:
- 我们必须通过取一个新的变量把给定的数字转换成字符串。
- 遍历字符串，将每个元素转换为整数，并将其相加。
- 再次遍历字符串
- 检查总和是否不能被任何一个数字整除
- 如果是真的，则返回假的
- 否则返回真
下面是上述方法的实现:
Python 3
# Python implementation of above approach
def getResult(n):
# Converting integer to string
st = str(n)
# Initialising sum to 0
sum = 0
length = len(st)
# Traversing through the string
for i in st:
# Converting character to int
sum = sum + int(i)
# Comparing number and sum
# Traversing again
for i in st:
# Check if any digit is
# not dividing the sum
if(sum % int(i) != 0):
# Return false
return 'No'
# If any value is not returned
# then all the digits are dividing the sum
# SO return true
return 'Yes'
# Driver Code
n = 123
# passing this number to get result function
print(getResult(n))
# this code is contributed by vikkycirus
java 描述语言
<script>
// JavaScript implementation of above approach
function getResult(n){
// Converting integer to string
var st = n.toString();
// Initialising sum to 0
var sum = 0
var length = st.length;
// Traversing through the string
for (var i= 0 ;i<st.length ; i++){
// Converting character to int
sum = sum + Number(st[i])
}
// Comparing number and sum
// Traversing again
for( var i = 0 ; i < st.length ; i++){
// Check if any digit is
// not dividing the sum
if(sum % Number(st[i]) != 0){
// Return false
return 'No'
}
// If any value is not returned
// then all the digits are dividing the sum
// SO return true
else{
return 'Yes';
}
}
}
// Driver Code
var n = 123;
// passing this number to get result function
document.write(getResult(n));
</script>
Output
Yes
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