public boolean checkPerfectNumber(intnum) {if(num<6||num%2!=0) {returnfalse; }intsum =1;for(inti=2; i<=num/2; i++) {if(num%i ==0) { sum += i; } }returnsum ==num; } 03 第二种解法 我们是不是可以将for循环中的循环次数再缩小一点?第一种解法是num/2次,而一个正整数它所...
The program output is also shown below. import java.util.Scanner; public class Perfect { public static void main(String[] args) { int n, sum = 0; Scanner s = new Scanner(System.in); System.out.print("Enter the number: "); n = s.nextInt(); for(int i = 1; i < n; i++...
按照你的要求编写的用于判断一个整数n是不是完数的Java程序如下 import java.util.Scanner;public class G { public static void perfectNumber(int n){ int f[]=new int[32]; int sum=0; int j=0; for(int i=1;i<n;i++){ if(n%i==0){ sum=sum+i; f[j]=i; ...
importjava.util.Scanner;publicclassPerfectNumber{publicstaticvoidmain(String[]args){Scannerscanner=newScanner(System.in);System.out.print("请输入一个正整数N:");intN=scanner.nextInt();scanner.close();for(inti=1;i<=N;i++){intsum=0;for(intj=1;j...
You can also usewhile loopto check the prime number: Just replace this part of the code in above program: for(inti=2;i<=num/2;i++){temp=num%i;if(temp==0){isPrime=false;break;}} with this: inti=2;while(i<=num/2){if(num%i==0){isPrime=false;break;}i++;} ...
According to number theory, a limb of pure mathematics that deals with integers, a Perfect Number can be defined as a positive integer whose value is
In the file Lagrange.java, write a Java program that takes a positive integer and breaks it into a sum of at most 4 perfect squares. At the heart of your program should be a method called findSum() that uses recursive backtracking to search for an appropriate sum. You will need to ...
while(number>0) { remdr=number%10; total=total+remdr; number=number/10; } System.out.print("Sum of digits of number "+tempNum+" is "+total); } } Output: Enter a Number : 5385 Sum of digits of number 5385 is 21 That’s all about Java program to add digits of number....
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2. Java Program to find deficient number publicclassMain { staticintdivsum(intn) { intsum =0; for(inti =1; i <= (Math.sqrt(n)); i++) { if(n % i ==0) { if(n / i == i) { sum = sum + i; }else{ sum = sum + i; ...