N 的最小除数 K 次重复相加后形成的数
原文:https://www . geesforgeks . org/number-formed-after-k-times-repeat-相加-n 的最小除数/
给定两个整数 N 和 K ，任务是生成执行 K 操作的最终结果，该操作包括在每一步将 N 的当前值的最小除数相加，而不是 1。 例:
输入: N = 9，K = 4 输出: 18 说明: 9 的除数为{1，3，9} 第 1 次运算:N = 9 + 3 = 12 第 2 次运算:N = 12 + 2 = 14 第 3 次运算:N = 14 + 2 = 16 第 4 次运算:N = 16 + 2 = 18 输入:【T11
天真法:这个问题的蛮力法是执行 K 次运算，然后打印最后一个数字。 高效方法:这里的诀窍是，如果给定的 N 是 甚至 ，则所有 K 运算的最小除数将始终为 2。因此，所需的第 K号将简单地为
所需数量= N + K * 2
还有如果 N 是奇数，最小除数将是奇数。因此，将它们相加将产生一个偶数值(奇数+奇数=偶数)。因此，现在上述技巧可以应用于(K-1)操作，即
所需数=N+N+的最小除数(K–1)* 2
以下是上述方法的实现:
C++
// C++ program to find the Kth number
// formed after repeated addition of
// smallest divisor of N
#include<bits/stdc++.h>
#include <cmath>
using namespace std;
void FindValue(int N, int K)
{
// If N is even
if( N % 2 == 0 )
{
N = N + 2 * K;
}
// If N is odd
else
{
int i;
for(i = 2; i < sqrt(N) + 1; i++)
{
if(N % i == 0)
break;
}
// Add smallest divisor to N
N = N + i;
// Updated N is even
N = N + 2 * ( K - 1 );
}
cout << N << endl;
}
// Driver code
int main()
{
int N = 9;
int K = 4;
FindValue( N, K );
}
// This code is contributed by Surendra_Gangwar
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the Kth number
// formed after repeated addition of
// smallest divisor of N
import java.util.*;
class GFG{
static void FindValue(int N, int K)
{
// If N is even
if( N % 2 == 0 )
{
N = N + 2 * K;
}
// If N is odd
else
{
int i;
for(i = 2; i < Math.sqrt(N) + 1; i++)
{
if(N % i == 0)
break;
}
// Add smallest divisor to N
N = N + i;
// Updated N is even
N = N + 2 * ( K - 1 );
}
System.out.print(N);
}
// Driver code
public static void main(String args[])
{
int N = 9;
int K = 4;
FindValue( N, K );
}
}
// This code is contributed by Nidhi_biet
Python 3
# Python3 program to find the
# Kth number formed after
# repeated addition of
# smallest divisor of N
import math
def FindValue(N, K):
# If N is even
if( N % 2 == 0 ):
N = N + 2 * K
# If N is odd
else:
# Find the smallest divisor
for i in range( 2, (int)(math.sqrt(N))+1 ):
if( N % i == 0):
break
# Add smallest divisor to N
N = N + i
# Updated N is even
N = N + 2 * ( K - 1 )
print(N)
# Driver code
if __name__ == "__main__":
N = 9
K = 4
FindValue( N, K )
C
// C# program to find the Kth number
// formed after repeated addition of
// smallest divisor of N
using System;
class GFG{
static void FindValue(int N, int K)
{
// If N is even
if( N % 2 == 0 )
{
N = N + 2 * K;
}
// If N is odd
else
{
int i;
for(i = 2; i < Math.Sqrt(N) + 1; i++)
{
if(N % i == 0)
break;
}
// Add smallest divisor to N
N = N + i;
// Updated N is even
N = N + 2 * ( K - 1 );
}
Console.WriteLine(N);
}
// Driver code
public static void Main()
{
int N = 9;
int K = 4;
FindValue( N, K );
}
}
// This code is contributed by Code_Mech
java 描述语言
<script>
// JavaScript program to find the Kth number
// formed after repeated addition of
// smallest divisor of N
function FindValue(N, K)
{
// If N is even
if( N % 2 == 0 )
{
N = N + 2 * K;
}
// If N is odd
else
{
let i;
for(i = 2; i < Math.sqrt(N) + 1; i++)
{
if(N % i == 0)
break;
}
// Add smallest divisor to N
N = N + i;
// Updated N is even
N = N + 2 * ( K - 1 );
}
document.write(N + "<br>");
}
// Driver code
let N = 9;
let K = 4;
FindValue( N, K );
// This code is contributed by Surbhi Tyagi.
</script>
Output:
18
时间复杂度:O(√N)T4】
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