检查一个数的奇数和偶数因子的计数是否相等
原文:https://www . geeksforgeeks . org/check-count-count-of-of-of-单数和双数因子是否相等/
给定一个数字 N ，任务是找出 N 是否有相等数量的奇偶因子。 例:
输入: N = 10 输出: YES 解释: 10 有两个奇数因子(1 和 5)和两个偶数因子(2 和 10) 输入: N = 24 输出: NO 解释: 24 有两个奇数因子(1 和 3)和六个偶数因子(2、4、6、8、12 和 24
天真的做法: 找出所有的除数检查奇数除数和偶数除数的计数是否相同。 以下是上述办法的实施情况
C++
// C++ solution for the above approach
#include <bits/stdc++.h>
using namespace std;
#define lli long long int
// Function to check condition
void isEqualFactors(lli N)
{
// Initialize even_count
// and od_count
lli ev_count = 0, od_count = 0;
// loop runs till square root
for (lli i = 1;
i <= sqrt(N) + 1; i++) {
if (N % i == 0) {
if (i == N / i) {
if (i % 2 == 0)
ev_count += 1;
else
od_count += 1;
}
else {
if (i % 2 == 0)
ev_count += 1;
else
od_count += 1;
if ((N / i) % 2 == 0)
ev_count += 1;
else
od_count += 1;
}
}
}
if (ev_count == od_count)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
// Driver Code
int main()
{
lli N = 10;
isEqualFactors(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java solution for the above approach
class GFG{
// Function to check condition
static void isEqualFactors(int N)
{
// Initialize even_count
// and od_count
int ev_count = 0, od_count = 0;
// Loop runs till square root
for(int i = 1;
i <= Math.sqrt(N) + 1; i++)
{
if (N % i == 0)
{
if (i == N / i)
{
if (i % 2 == 0)
ev_count += 1;
else
od_count += 1;
}
else
{
if (i % 2 == 0)
ev_count += 1;
else
od_count += 1;
if ((N / i) % 2 == 0)
ev_count += 1;
else
od_count += 1;
}
}
}
if (ev_count == od_count)
System.out.print("YES" + "\n");
else
System.out.print("NO" + "\n");
}
// Driver Code
public static void main(String[] args)
{
int N = 10;
isEqualFactors(N);
}
}
// This code is contributed by amal kumar choubey
Python 3
# Python3 code for the naive approach
import math
# Function to check condition
def isEqualFactors(N):
# Initialize even_count
# and od_count
ev_count = 0
od_count = 0
# loop runs till square root
for i in range(1, (int)(math.sqrt(N)) + 2) :
if (N % i == 0):
if(i == N // i):
if(i % 2 == 0):
ev_count += 1
else:
od_count += 1
else:
if(i % 2 == 0):
ev_count += 1
else:
od_count += 1
if((N // i) % 2 == 0):
ev_count += 1;
else:
od_count += 1;
if (ev_count == od_count):
print("YES")
else:
print("NO")
# Driver Code
N = 10
isEqualFactors(N)
C
// C# solution for the above approach
using System;
class GFG{
// Function to check condition
static void isEqualFactors(int N)
{
// Initialize even_count
// and od_count
int ev_count = 0, od_count = 0;
// Loop runs till square root
for(int i = 1;
i <= Math.Sqrt(N) + 1; i++)
{
if (N % i == 0)
{
if (i == N / i)
{
if (i % 2 == 0)
ev_count += 1;
else
od_count += 1;
}
else
{
if (i % 2 == 0)
ev_count += 1;
else
od_count += 1;
if ((N / i) % 2 == 0)
ev_count += 1;
else
od_count += 1;
}
}
}
if (ev_count == od_count)
Console.Write("YES" + "\n");
else
Console.Write("NO" + "\n");
}
// Driver Code
public static void Main(String[] args)
{
int N = 10;
isEqualFactors(N);
}
}
// This code is contributed by gauravrajput1
java 描述语言
<script>
// Javascript program for the above approach
// Function to check condition
function isEqualFactors(N)
{
// Initialize even_count
// and od_count
let ev_count = 0, od_count = 0;
// Loop runs till square root
for(let i = 1;
i <= Math.sqrt(N) + 1; i++)
{
if (N % i == 0)
{
if (i == N / i)
{
if (i % 2 == 0)
ev_count += 1;
else
od_count += 1;
}
else
{
if (i % 2 == 0)
ev_count += 1;
else
od_count += 1;
if ((N / i) % 2 == 0)
ev_count += 1;
else
od_count += 1;
}
}
}
if (ev_count == od_count)
document.write("YES" + "\n");
else
document.write("NO" + "\n");
}
// Driver Code
let N = 10;
isEqualFactors(N);
</script>
Output:
YES
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