检查所有数组元素串联形成的数字是否为 Harshad 数
原文:https://www . geesforgeks . org/check-如果通过串联所有数组元素形成的数字是一个-harshad-number-or-not/
给定一个由 N 个整数组成的数组arr【】，任务是检查通过连接所有数组元素形成的数字是否为 Harshad 数字。
示例:
输入 : arr[] = { 1，35，69，60} 输出:是 说明: 数组元素串联形成的个数为“1356960”。 数字的位数之和= 1 + 3 + 5 + 6 + 9 + 6 + 0 = 30。 由于数字可被其位数的和整除，所以该数字是“哈沙德数”。
输入: arr[] = {1，563，9，59，7，8 } T3】输出:是
方法:想法是将所有数组元素转换为它们的等价字符串并连接这些字符串。按照以下步骤解决问题:
- 遍历数组 arr[] ，并将每个数组元素转换为其等价字符串。
- 连接变量中的所有字符串，比如 S.
- 初始化一个变量，说和，存储生成数的位数和。
- 遍历字符串 S 并将 sum 更新为 sum += int (s[i])。
- 初始化一个变量，比如说 N = 0，来存储由字符串 S mod sum 的所有字符连接而成的数字。
- 遍历字符串 S 并将 N 更新为 N = (N * 10 + int (S[i]))的% sum。
- 打印是，如果 N = 0。
- 否则，打印否。
下面是上述方法的实现:
C++
// CPP implementation
// of above approach
#include<bits/stdc++.h>
using namespace std;
// Function to check whether n
// is a harshad number or not
int checkHarshad(string n)
{
// Stores the sum of digits
int sum = 0;
// Stores the remainder
int N = 0;
// Increment sum
for (int c = 0; c < n.length(); c++)
sum += (n);
for (int c = 0; c < n.length(); c++)
{
N = N + (N * 10 + (n));
N %= sum;
}
return (N != 0);
}
// Function to check if concatenation
// of elements from the array arr[]
// is a harshad number or not
bool combineArray(vector<int> lis)
{
// Stores the concatenated number
string st = "";
// Traverse the array
for(auto el: lis)
{
// Concatenate the string
st += to_string(el);
}
if(checkHarshad(st))
return true;
else
return false;
}
// Driver Code
int main()
{
// Input
vector<int>arr{1, 35, 69, 60};
// Function call to check if
// concatenation of elements of
// arr[] is a harshad number
if(combineArray(arr))
cout << "Yes";
else
cout << "No";
}
// This code is contributed by ipg2016107.
Java 语言(一种计算机语言，尤用于创建网站)
/*package whatever //do not write package name here */
import java.io.*;
import java.util.ArrayList;
class GFG {
// Function to check whether n
// is a harshad number or not
public static boolean checkHarshad(String n)
{
// Stores the sum of digits
int sum = 0;
// Stores the remainder
int N = 0;
// Increment sum
for (int c = 0; c < n.length(); c++) {
sum += Character.getNumericValue(n.charAt(c));
}
for (int c = 0; c < n.length(); c++) {
N = N
+ (N * 10
+ (Character.getNumericValue(
n.charAt(c))));
N %= sum;
}
if (N == 0) {
return true;
}
return false;
}
// Function to check if concatenation
// of elements from the array arr[]
// is a harshad number or not
public static boolean
combineArray(ArrayList<Integer> list)
{
// Stores the concatenated number
String st = "";
// Traverse the array
for (int i = 0; i < list.size(); i++)
{
// Concatenate the string
st += Integer.toString(list.get(i));
}
if (checkHarshad(st)) {
return true;
}
return false;
}
// Driver Code
public static void main(String[] args)
{
// Input
ArrayList<Integer> list = new ArrayList<>();
list.add(1);
list.add(35);
list.add(69);
list.add(60);
// Function call to check if
// concatenation of elements of
// arr[] is a harshad number
if (combineArray(list))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by aditya7409
Python 3
# Python implementation
# of above approach
# Function to check whether n
# is a harshad number or not
def checkHarshad(n):
# Stores the sum of digits
sum = 0
# Stores the remainder
N = 0
# Increment sum
for c in n:
sum += int(c)
for c in n:
N = N + (N*10+int(c))
N %= sum
return N == 0
# Function to check if concatenation
# of elements from the array arr[]
# is a harshad number or not
def combineArray(lis):
# Stores the concatenated number
string=""
# Traverse the array
for el in lis:
# Convert to equivalent string
el = str(el)
# Concatenate the string
string = string + el
if(checkHarshad(string)):
return True
else:
return False
# Driver Code
# Input
arr=[1, 35, 69, 60]
# Function call to check if
# concatenation of elements of
# arr[] is a harshad number
if(combineArray(arr)):
print("Yes")
else:
print("No")
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to check whether n
// is a harshad number or not
public static bool checkHarshad(string n)
{
// Stores the sum of digits
int sum = 0;
// Stores the remainder
int N = 0;
// Increment sum
for (int c = 0; c < n.Length; c++) {
sum += (int)Char.GetNumericValue(n);
}
for (int c = 0; c < n.Length; c++) {
N = N
+ (N * 10
+ (int)(Char.GetNumericValue(
n)));
N %= sum;
}
if (N == 0) {
return true;
}
return false;
}
// Function to check if concatenation
// of elements from the array arr[]
// is a harshad number or not
static bool
combineArray(List<int> list)
{
// Stores the concatenated number
string st = "";
st += string.Join("", list);
if (checkHarshad(st)) {
return true;
}
return false;
}
// Driver code
static void Main()
{
List<int> list = new List<int>();
list.Add(1);
list.Add(35);
list.Add(69);
list.Add(60);
// Function call to check if
// concatenation of elements of
// arr[] is a harshad number
if (combineArray(list))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed susmitakundugoaldanga.
java 描述语言
<script>
// Javascript implementation
// of above approach
// Function to check whether n
// is a harshad number or not
function checkHarshad(n)
{
// Stores the sum of digits
var sum = 0;
// Stores the remainder
var N = 0;
// Increment sum
for (var c = 0; c < n.length; c++)
sum += (n);
for (var c = 0; c < n.length; c++)
{
N = N + (N * 10 + (n));
N %= sum;
}
return (N != 0);
}
// Function to check if concatenation
// of elements from the array arr[]
// is a harshad number or not
function combineArray(lis)
{
// Stores the concatenated number
var st = "";
// Traverse the array
for(var el in lis)
{
// Concatenate the string
st += (el.toString());
}
if(checkHarshad(st))
return true;
else
return false;
}
// Driver Code
// Input
var arr = [1, 35, 69, 60];
// Function call to check if
// concatenation of elements of
// arr[] is a harshad number
if(combineArray(arr))
document.write( "Yes");
else
document.write( "No");
</script>
Output:
Yes
时间复杂度:O(N) T5辅助空间:** O(N)
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