连接编号
结号是一个数字 N 如果它可以写成 K + SumOfDIgits(k)至少两个 K 的话
101, 103, 105, 107, 109, 111, 113, 115….
检查数字 N 是否为交叉点编号
给定一个数字 N ，任务是检查 N 是否为路口号。如果 N 是一个连接编号，则打印“是”否则打印“否”。 示例:
输入: N = 1106 输出:是 说明: 1106 = 1093+(1+0+9+3)= 1102+(1+1+0+2) 输入: N = 11 输出:否
方法:由于结数是一个数 N 如果它可以写成 K + SumOfDIgits(k)至少两个 K，那么对于一个循环中从 1 到 N 的所有数，我们将检查 I+SumOfDIgits(I)是否等于 N。如果等于 N 打印“是”否则打印“否”。 以下是上述方法的实施:
C++
// C++ implementation for the
// above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the
// sum Of digits of N
int sum(int n)
{
// To store sum of N and
// sumOfdigitsof(N)
int sum = 0;
while (n != 0) {
// extracting digit
int r = n % 10;
sum = sum + r;
n = n / 10;
}
return sum;
}
// Function to check Junction numbers
bool isJunction(int n)
{
// To store count of ways n can be
// represented as i + SOD(i)
int count = 0;
for (int i = 1; i <= n; i++) {
if (i + sum(i) == n)
count++;
}
return count >= 2;
}
// Driver Code
int main()
{
int N = 111;
// Function Call
if (isJunction(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 program for above approach
class GFG{
// Function to find the
// sum Of digits of N
static int sum(int n)
{
// To store sum of N and
// sumOfdigitsof(N)
int sum = 0;
while (n != 0)
{
// extracting digit
int r = n % 10;
sum = sum + r;
n = n / 10;
}
return sum;
}
// Function to check Junction numbers
static boolean isJunction(int n)
{
// To store count of ways n can be
// represented as i + SOD(i)
int count = 0;
for (int i = 1; i <= n; i++)
{
if (i + sum(i) == n)
count++;
}
return count >= 2;
}
// Driver Code
public static void main(String[] args)
{
int N = 111;
// Function Call
if (isJunction(N))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by Pratima Pandey
Python 3
# Python3 program for the above approach
import math
# Function to find the
# sum Of digits of N
def sum1(n):
# To store sum of N and
# sumOfdigitsof(N)
sum1 = 0
while (n != 0):
# extracting digit
r = n % 10
sum1 = sum1 + r
n = n // 10
return sum1
# Function to check Junction numbers
def isJunction(n):
# To store count of ways n can be
# represented as i + SOD(i)
count = 0
for i in range(1, n + 1):
if (i + sum1(i) == n):
count = count + 1
return count >= 2
# Driver Code
if __name__=='__main__':
# Given Number
n = 111
# Function Call
if (isJunction(n) == 1):
print("Yes")
else:
print("No")
# This code is contributed by rock_cool
C
// C# program for above approach
using System;
class GFG{
// Function to find the
// sum Of digits of N
static int sum(int n)
{
// To store sum of N and
// sumOfdigitsof(N)
int sum = 0;
while (n != 0)
{
// extracting digit
int r = n % 10;
sum = sum + r;
n = n / 10;
}
return sum;
}
// Function to check Junction numbers
static bool isJunction(int n)
{
// To store count of ways n can be
// represented as i + SOD(i)
int count = 0;
for (int i = 1; i <= n; i++)
{
if (i + sum(i) == n)
count++;
}
return count >= 2;
}
// Driver Code
public static void Main()
{
int N = 111;
// Function Call
if (isJunction(N))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Nidhi Biet
java 描述语言
<script>
// Javascript program for above approach
// Function to find the
// sum Of digits of N
function sum( n) {
// To store sum of N and
// sumOfdigitsof(N)
let sum = 0;
while (n != 0) {
// extracting digit
let r = n % 10;
sum = sum + r;
n = parseInt(n / 10);
}
return sum;
}
// Function to check Junction numbers
function isJunction( n) {
// To store count of ways n can be
// represented as i + SOD(i)
let count = 0;
for ( i = 1; i <= n; i++) {
if (i + sum(i) == n)
count++;
}
return count >= 2;
}
// Driver Code
let N = 111;
// Function Call
if (isJunction(N))
document.write("Yes");
else
document.write("No");
// This code contributed by Rajput-Ji
</script>
Output:
Yes
时间复杂度:O(n) T5】参考:http://www.numbersaplenty.com/set/junction_number/
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