检查给定数字是否为完全数的 Java 程序
如果一个数的适当除数之和(即不包括数本身的所有正除数)等于该数本身，则称该数为。等分总和是一个数的除数之和，不包括数本身。因此，一个数只有等于它的等份和时才是完全数。所有已知的完全数都是甚至。
**Example 1:**
n = 9
Proper Divisors of 9 are 1 and 3.
Sum = 1+3 = 4 ≠ 9
⇒ 9 is not a perfect number.
**Example 2:**
n = 6
Proper Divisors of 6 are 1, 2 and 3.
Sum = 1+2+3 = 6 = 6
⇒ 6 is a perfect number.
**Example 3:**
n = 28
Proper Divisors of 28 are 1, 2, 4, 7 and 14.
Sum = 1+2+4+7+14 = 28 = 28
⇒ 28 is a perfect number.
**Example 4:**
n = 15
Proper Divisors of 15 are 1,3 and 5.
Sum = 1+3+5 = 9 ≠ 15
⇒ 15 is not a perfect number.
所以，我们基本上要找到一个数的适当除数的和。
*方法 1:*
一个简单的解决方法就是把从 1 到 n-1 的每一个数都过一遍，检查它是不是除数，如果是，那么把它加到和变量中，最后检查和是否等于数本身，那么它就是一个完全数，否则不是。
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to check if a given
// number is perfect or not
class GFG {
// Returns true if n is perfect
static boolean isPerfect(int n)
{
// 1 is not a perfect number
if (n == 1)
return false;
// sum will store the sum of proper divisors
// As 1 is a proper divisor for all numbers
// initialised sum with 1
int sum = 1;
// Looping through the numbers to check if they are
// divisors or not
for (int i = 2; i < n; i++) {
if (n % i == 0) {
sum += i;
}
}
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum == n)
return true;
return false;
}
// Driver program
public static void main(String[] args)
{
int n = 6;
// Call isPerfect function to
// check if the number is perfect or not.
if (isPerfect(n))
System.out.println(n + " is a perfect number");
else
System.out.println(
n + " is not a perfect number");
}
}
**Output
6 is a perfect number
```**
* ****时间复杂度:** O(n)**
****方法 2:****
**一个**有效的解决方案**是遍历数字直到 n 的**平方根****
> **如果 **i** 是除数，那么 **n/i** 也是除数。**
## **Java**
```java
// Java program to check if a given
// number is perfect or not
class GFG {
// Returns true if n is perfect
static boolean isPerfect(int n)
{
// 1 is not a perfect number
if (n == 1)
return false;
// sum will store the sum of proper divisors
// As 1 is a proper divisor for all numbers
// initialised sum with 1
int sum = 1;
// Looping through the numbers to check if they are
// divisors or not
for (int i = 2; i * i <= n; i++) {
if (n % i == 0) {
// n is a perfect square
// let's take 25
// we need to add 5 only once
// sum += i + n / i will add it twice
if (i * i == n)
sum += i;
else
sum += i + (n / i);
}
}
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum == n)
return true;
return false;
}
// Driver program
public static void main(String[] args)
{
int n = 6;
// Call isPerfect function to
// check if the number is perfect or not.
if (isPerfect(n))
System.out.println(n + " is a perfect number");
else
System.out.println(
n + " is not a perfect number");
}
}
**输出
java
6 is a perfect number**
*时间复杂度:* O(√n)
版权属于：月萌API www.moonapi.com，转载请注明出处
