目瞪口呆的数字
间隙数是一个至少有 3 位数字的数 N ,这样它可以被它的第一个和最后一个数字的连接整除。 很少有空白的数字是:
100、105、108、110、120、121、130、132、135、140……
检查 N 是否为空白数
给定一个整数 N ,任务是检查 N 是否为间隙数。如果 N 是空白号码,则打印“是”否则打印“否”。 举例:
输入: N = 108 输出:是 说明: 108 可被 18 整除 输入: N = 112 输出:否
方法:思路是用给定数字的首末数字创建一个数字(比如 num ,检查 N 是否可以被 num 整除。如果 N 可以被 num 整除,那么它就是一个间隙数,打印“是”,否则打印“否”。 以下是上述方法的实施:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Find the first digit
int firstDigit(int n)
{
// Find total number of digits - 1
int digits = (int)log10(n);
// Find first digit
n = (int)(n / pow(10, digits));
// Return first digit
return n;
}
// Find the last digit
int lastDigit(int n)
{
// return the last digit
return (n % 10);
}
// A function to check Gapful numbers
bool isGapful(int n)
{
int first_dig = firstDigit(n);
int last_dig = lastDigit(n);
int concatenation = first_dig * 10
+ last_dig;
// Return true if n is gapful number
return (n % concatenation == 0);
}
// Driver Code
int main()
{
// Given Number
int n = 108;
// Function Call
if (isGapful(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 the above approach
class GFG{
// Find the first digit
static int firstDigit(int n)
{
// Find total number of digits - 1
int digits = (int)(Math.log(n) /
Math.log(10));
// Find first digit
n = (int)(n / Math.pow(10, digits));
// Return first digit
return n;
}
// Find the last digit
static int lastDigit(int n)
{
// Return the last digit
return (n % 10);
}
// A function to check Gapful numbers
static boolean isGapful(int n)
{
int first_dig = firstDigit(n);
int last_dig = lastDigit(n);
int concatenation = first_dig * 10 +
last_dig;
// Return true if n is gapful number
return (n % concatenation == 0);
}
// Driver code
public static void main(String[] args)
{
// Given number
int n = 108;
// Function call
if (isGapful(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
# Find the first digit
def firstDigit(n):
# Find total number of digits - 1
digits = math.log10(n)
# Find first digit
n = (n / math.pow(10, digits))
# Return first digit
return n
# Find the last digit
def lastDigit(n):
# return the last digit
return (n % 10)
# A function to check Gapful numbers
def isGapful(n):
concatenation = (firstDigit(n) * 10) +\
lastDigit(n)
# Return true if n is gapful number
return (n % concatenation)
# Driver Code
if __name__=='__main__':
# Given Number
n = 108
# Function Call
if (isGapful(n)):
print("Yes")
else:
print("No")
# This code is contributed by Ritik Bansal
C
// C# program for the above approach
using System;
class GFG{
// Find the first digit
static int firstDigit(int n)
{
// Find total number of digits - 1
int digits = (int)(Math.Log(n) /
Math.Log(10));
// Find first digit
n = (int)(n / Math.Pow(10, digits));
// Return first digit
return n;
}
// Find the last digit
static int lastDigit(int n)
{
// Return the last digit
return (n % 10);
}
// A function to check Gapful numbers
static bool isGapful(int n)
{
int first_dig = firstDigit(n);
int last_dig = lastDigit(n);
int concatenation = first_dig * 10 +
last_dig;
// Return true if n is gapful number
return (n % concatenation == 0);
}
// Driver code
public static void Main()
{
// Given number
int n = 108;
// Function call
if (isGapful(n))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Code_Mech
java 描述语言
<script>
// Javascript program for the above approach
// Find the first digit
function firstDigit( n)
{
// Find total number of digits - 1
let digits = parseInt( (Math.log(n) / Math.log(10)));
// Find first digit
n = parseInt( (n / Math.pow(10, digits)));
// Return first digit
return n;
}
// Find the last digit
function lastDigit( n)
{
// Return the last digit
return (n % 10);
}
// A function to check Gapful numbers
function isGapful( n)
{
let first_dig = firstDigit(n);
let last_dig = lastDigit(n);
let concatenation = first_dig * 10 + last_dig;
// Return true if n is gapful number
return (n % concatenation == 0);
}
// Driver code
// Given number
let n = 108;
// Function call
if (isGapful(n))
document.write("Yes");
else
document.write("No");
// This code is contributed by aashish1995
</script>
Output:
Yes
时间复杂度:O(1) T5】参考:https://oeis.org/A108343
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