求不能被 A 整除的第 n 个自然数
给定两个整数 A 和 N ，我们的任务是找到不能被 A 整除的第 N 个自然数。
示例:
输入: A = 4，N = 12 输出: 15 说明: 从 1 开始除 A 的倍数外的数列为 1、2、3、5、6、7、9、10、11、13、14、15、17，不能被 4 整除的第 12 项为 15。
输入: A = 3，N = 20 输出: 29 说明: 除 A 的倍数外，从 1 开始的数列为 1、2、4、5、7、8、10、11 等，不能被 3 整除的第 N 个数为 29。
进场:
为了解决上面提到的问题，我们必须观察到每(A–1)个整数后都有一个间隙，因为跳过了 A 的倍数。为了发现数字 N 位于哪个集合中，我们将 N 除以(A–1)，并将其存储在一个变量中，让我们说商。现在将 A 与那个变量相乘，并将余数相加，我们将得到结果。
If A = 4, N = 7:
quotient = 7 / 3 = 2
remainder = 7 % 3 = 1
So the answer is
(A * quotient) + remainder
= 4 * 2 + 1 = 9
但是如果余数为 0，则意味着它是集合中的最后一个元素。让我们通过一个例子来理解它。
If A = 4, N = 6:
quotient = 6 / 3 = 2
remainder = 6 % 3 = 0
So the answer is
(A * quotient) - 1
= 4 * 2 - 1 = 7
下面是上述方法的实现:
C++
// C++ code to Find the Nth number which
// is not divisible by A from the series
#include<bits/stdc++.h>
using namespace std;
void findNum(int n, int k)
{
// Find the quotient and the remainder
// when k is divided by n-1
int q = k / (n - 1);
int r = k % (n - 1);
int a;
// If the remainder is not 0
// multiply n by q and subtract 1
// if remainder is 0 then
// multiply n by q
// and add the remainder
if(r != 0)
a = (n * q) + r;
else
a = (n * q) - 1;
// Print the answer
cout << a;
}
// Driver code
int main()
{
int A = 4, N = 6;
findNum(A, N);
return 0;
}
// This code is contributed by PratikBasu
Java 语言(一种计算机语言，尤用于创建网站)
// Java code to Find the Nth number which
// is not divisible by A from the series
class GFG {
static void findNum(int n, int k)
{
// Find the quotient and the remainder
// when k is divided by n-1
int q = k / (n - 1);
int r = k % (n - 1);
int a = 0;
// If the remainder is not 0
// multiply n by q and subtract 1
// if remainder is 0 then
// multiply n by q
// and add the remainder
if (r != 0)
a = (n * q) + r;
else
a = (n * q) - 1;
// Print the answer
System.out.println(a);
}
// Driver code
public static void main(String[] args)
{
int A = 4;
int N = 6;
findNum(A, N);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 code to Find the Nth number which
# is not divisible by A from the series
def findNum(n, k):
# Find the quotient and the remainder
# when k is divided by n-1
q = k//(n-1)
r = k % (n-1)
# if the remainder is not 0
# multiply n by q and subtract 1
# if remainder is 0 then
# multiply n by q
# and add the remainder
if(r != 0):
a = (n * q)+r
else:
a = (n * q)-1
# print the answer
print(a)
# driver code
A = 4
N = 6
findNum(A, N)
C
// C# code to find the Nth number which
// is not divisible by A from the series
using System;
class GFG{
static void findNum(int n, int k)
{
// Find the quotient and the remainder
// when k is divided by n-1
int q = k / (n - 1);
int r = k % (n - 1);
int a = 0;
// If the remainder is not 0
// multiply n by q and subtract 1
// if remainder is 0 then
// multiply n by q
// and add the remainder
if (r != 0)
a = (n * q) + r;
else
a = (n * q) - 1;
// Print the answer
Console.WriteLine(a);
}
// Driver code
public static void Main(String[] args)
{
int A = 4;
int N = 6;
findNum(A, N);
}
}
// This code is contributed by amal kumar choubey
java 描述语言
<script>
// Javascript code to Find the Nth number which
// is not divisible by A from the series
function findNum(n, k)
{
// Find the quotient and the remainder
// when k is divided by n-1
let q = parseInt(k / (n - 1));
let r = k % (n - 1);
let a;
// If the remainder is not 0
// multiply n by q and subtract 1
// if remainder is 0 then
// multiply n by q
// and add the remainder
if (r != 0)
a = (n * q) + r;
else
a = (n * q) - 1;
// Print the answer
document.write(a);
}
// Driver code
let A = 4, N = 6;
findNum(A, N);
// This code is contributed by rishavmahato348
</script>
Output:
7
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