LAB1.docx
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This document contains 6 Java programs that demonstrate basic programming concepts like input/output, conditionals, loops, and functions. Program 1 takes input from the user and prints the sum. Program 2 prints numbers in a range. Program 3 finds the largest of 3 numbers. Program 4 checks if a number is prime. Program 5 prints the Fibonacci series up to 10 terms. Each program shows the code and expected output.
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![JAVA
LAB 1
1:
CODE:-
import java.util.Scanner;
public class Main
{
public static void main(String[] args) {
int n1,n2;
Scanner op=new Scanner(System.in);
System.out.println("Enter No:");
n1 = op.nextInt();
n2 = op.nextInt();
System.out.println(" Sum of two no:"+(n1+n2));
}
}](https://image.slidesharecdn.com/lab1-220506192621-4fe00cf9/85/LAB1-docx-1-320.jpg)
![OUTPUT:-
3:
CODE:
import java.util.Scanner;
public class Main
{
public static void main(String[] args) {
int l,u;
Scanner op=new Scanner(System.in);
System.out.println("Enter lower limit and upper
limit :");
l = op.nextInt();
u = op.nextInt();
System.out.println("No. in increasing order:");
for(int i=l+1;i<=u-1;i++)
{
System.out.printf("%d ",i);](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
Codeimport java.util.Random; import java.util.Scanner;public .pdf
Code:
import java.util.Random;
import java.util.Scanner;
public class Test{
public static void main(String args[]){
Scanner input = new Scanner(System.in);
String bin = \"1\";
int n;
System.out.print(\"Enter the number of binary digits : \");
n = input.nextInt();
for(int i = 0; i < n-1; i++){
bin += randomBin();
}
System.out.println(bin);
}
public static String randomBin() {
Random rg = new Random();
int n = rg.nextInt(2);
return Integer.toBinaryString(n);
}
}
Output:
Enter the number of binary digits : 5
10110
Solution
Code:
import java.util.Random;
import java.util.Scanner;
public class Test{
public static void main(String args[]){
Scanner input = new Scanner(System.in);
String bin = \"1\";
int n;
System.out.print(\"Enter the number of binary digits : \");
n = input.nextInt();
for(int i = 0; i < n-1; i++){
bin += randomBin();
}
System.out.println(bin);
}
public static String randomBin() {
Random rg = new Random();
int n = rg.nextInt(2);
return Integer.toBinaryString(n);
}
}
Output:
Enter the number of binary digits : 5
10110.
PrimeRange.java import java.util.Scanner;public class PrimeRan.pdf
PrimeRange.java
import java.util.Scanner;
public class PrimeRange {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter low range:\");
int low = scan.nextInt();
System.out.println(\"Enter high range:\");
int high = scan.nextInt();
for(int i=low; i<=high; i++){
if(isPrime(i))
System.out.print(i+\" \");
}
}
public static boolean isPrime(int n){
for (int f = 2; f < n / 2; f++) {
if (n % f == 0) {
return false;
}
}
return true;
}
}
Output:
Enter low range:
10
Enter high range:
30
11 13 17 19 23 29
Solution
PrimeRange.java
import java.util.Scanner;
public class PrimeRange {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter low range:\");
int low = scan.nextInt();
System.out.println(\"Enter high range:\");
int high = scan.nextInt();
for(int i=low; i<=high; i++){
if(isPrime(i))
System.out.print(i+\" \");
}
}
public static boolean isPrime(int n){
for (int f = 2; f < n / 2; f++) {
if (n % f == 0) {
return false;
}
}
return true;
}
}
Output:
Enter low range:
10
Enter high range:
30
11 13 17 19 23 29.
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}
Solution
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}.
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}
Solution
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}.
Recommended
Codeimport java.util.Random; import java.util.Scanner;public .pdf
Code:
import java.util.Random;
import java.util.Scanner;
public class Test{
public static void main(String args[]){
Scanner input = new Scanner(System.in);
String bin = \"1\";
int n;
System.out.print(\"Enter the number of binary digits : \");
n = input.nextInt();
for(int i = 0; i < n-1; i++){
bin += randomBin();
}
System.out.println(bin);
}
public static String randomBin() {
Random rg = new Random();
int n = rg.nextInt(2);
return Integer.toBinaryString(n);
}
}
Output:
Enter the number of binary digits : 5
10110
Solution
Code:
import java.util.Random;
import java.util.Scanner;
public class Test{
public static void main(String args[]){
Scanner input = new Scanner(System.in);
String bin = \"1\";
int n;
System.out.print(\"Enter the number of binary digits : \");
n = input.nextInt();
for(int i = 0; i < n-1; i++){
bin += randomBin();
}
System.out.println(bin);
}
public static String randomBin() {
Random rg = new Random();
int n = rg.nextInt(2);
return Integer.toBinaryString(n);
}
}
Output:
Enter the number of binary digits : 5
10110.
PrimeRange.java import java.util.Scanner;public class PrimeRan.pdf
PrimeRange.java
import java.util.Scanner;
public class PrimeRange {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter low range:\");
int low = scan.nextInt();
System.out.println(\"Enter high range:\");
int high = scan.nextInt();
for(int i=low; i<=high; i++){
if(isPrime(i))
System.out.print(i+\" \");
}
}
public static boolean isPrime(int n){
for (int f = 2; f < n / 2; f++) {
if (n % f == 0) {
return false;
}
}
return true;
}
}
Output:
Enter low range:
10
Enter high range:
30
11 13 17 19 23 29
Solution
PrimeRange.java
import java.util.Scanner;
public class PrimeRange {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter low range:\");
int low = scan.nextInt();
System.out.println(\"Enter high range:\");
int high = scan.nextInt();
for(int i=low; i<=high; i++){
if(isPrime(i))
System.out.print(i+\" \");
}
}
public static boolean isPrime(int n){
for (int f = 2; f < n / 2; f++) {
if (n % f == 0) {
return false;
}
}
return true;
}
}
Output:
Enter low range:
10
Enter high range:
30
11 13 17 19 23 29.
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}
Solution
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}.
The sample program for above series in JAVA will be like belowimpo.pdf
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}
Solution
The sample program for above series in JAVA will be like below
import java.util.Scan;
public class PIVALUE {
public static void main(String[] args) {
// Scan object creation
Scan input = new Scan (System.in);
// ask user for input
System.out.println(\"Enter num.\");
double j = input.nextDouble(); // user entered value of j
double Tot = 0;
for(j=0; j<10000; j++){
if(j%2 == 0) // check remainder of `j/2` is 0
Tot += -1 / ( 2 * j - 1);
else
Tot += 1 / (2 * j - 1);
}
System.out.println(Tot);
}
}.
Programs of java
some programs of java language for more info contact 03023688542 Email: shafique.sangi786@gmail.com
CODEimport java.util.; public class test { public static voi.pdf
CODE:
import java.util.*;
public class test {
public static void main(String args[]) {
int num, sum=0;
Scanner input=new Scanner(System.in);
System.out.print(\"Input num : \");
num=input.nextInt();
while(num != 13) {
sum=sum+num;
System.out.print(\"Input num : \");
num=input.nextInt();
}
sum=sum+num;
System.out.println(\"Sum = \"+sum);
}
}
save the above file as test.java
Output:
javac test.java
java test
Input num :
1
Input num :
2
Input num :
3
Input num :
4
Input num :
5
Input num :
6
Input num :
7
Input num :
8
Input num :
9
Input num :
0
Input num :
13
Sum = 58
Solution
CODE:
import java.util.*;
public class test {
public static void main(String args[]) {
int num, sum=0;
Scanner input=new Scanner(System.in);
System.out.print(\"Input num : \");
num=input.nextInt();
while(num != 13) {
sum=sum+num;
System.out.print(\"Input num : \");
num=input.nextInt();
}
sum=sum+num;
System.out.println(\"Sum = \"+sum);
}
}
save the above file as test.java
Output:
javac test.java
java test
Input num :
1
Input num :
2
Input num :
3
Input num :
4
Input num :
5
Input num :
6
Input num :
7
Input num :
8
Input num :
9
Input num :
0
Input num :
13
Sum = 58.
import java.util.;public class Program{public static void.pdf
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
Solution
Hi,
I have modified the code. it is working as expected now. Highlighted the code changes below.
Issue here is with you have written this statement twice System.out.println(x); which causes the
issue.
Program.java
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println();
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Output:
Enter amount
10
Amount is: 10
X
XX
XXX
XXXX
XXXXX
XXXXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
XXXXXXXXXX.
Java programs
Sample java programs for beginners. This slides can help you to print few words in java. Also it helps to do arithmetic calculations and will give you basic concept of for loop, do while loop and if else statements in java. Area of a square.
SumNumbers.java import java.util.Scanner;public class SumNumbe.pdf
SumNumbers.java
import java.util.Scanner;
public class SumNumbers {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
int sum = 0;
while(true){
System.out.println(\"Enter integer:\");
int n = scan.nextInt();
if(n == -1){
break;
}
sum = sum + n;
}
System.out.println(\"The sum is \"+sum);
}
}
Output:
Enter integer:
1
Enter integer:
2
Enter integer:
3
Enter integer:
4
Enter integer:
5
Enter integer:
6
Enter integer:
7
Enter integer:
8
Enter integer:
9
Enter integer:
-1
The sum is 45
Solution
SumNumbers.java
import java.util.Scanner;
public class SumNumbers {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
int sum = 0;
while(true){
System.out.println(\"Enter integer:\");
int n = scan.nextInt();
if(n == -1){
break;
}
sum = sum + n;
}
System.out.println(\"The sum is \"+sum);
}
}
Output:
Enter integer:
1
Enter integer:
2
Enter integer:
3
Enter integer:
4
Enter integer:
5
Enter integer:
6
Enter integer:
7
Enter integer:
8
Enter integer:
9
Enter integer:
-1
The sum is 45.
Codeimport java.util.; class Test{ public static void main(S.pdf
Code:
import java.util.*;
class Test{
public static void main(String args[])throws Exception{
Scanner input = new Scanner(System.in);
System.out.print(\"Enter the First Number : \");
int n1=input.nextInt();
System.out.print(\"Enter the Second Number : \");
int n2=input.nextInt();
System.out.println(\"GCD : \"+gcd(n1,n2));
}
public static int gcd(int n1,int n2){
int r;
while(n2 != 0){
r = n1 % n2;
n1 = n2;
n2 = r;
}
return n1;
}
}
Output:
Enter the First Number : 54
Enter the Second Number : 24
GCD : 6
Solution
Code:
import java.util.*;
class Test{
public static void main(String args[])throws Exception{
Scanner input = new Scanner(System.in);
System.out.print(\"Enter the First Number : \");
int n1=input.nextInt();
System.out.print(\"Enter the Second Number : \");
int n2=input.nextInt();
System.out.println(\"GCD : \"+gcd(n1,n2));
}
public static int gcd(int n1,int n2){
int r;
while(n2 != 0){
r = n1 % n2;
n1 = n2;
n2 = r;
}
return n1;
}
}
Output:
Enter the First Number : 54
Enter the Second Number : 24
GCD : 6.
import java.util.Scanner; public class Palin { public static v.pdf
import java.util.Scanner;
public class Palin
{
public static void main(String args[])
{
Scanner s=new Scanner(System.in);
System.out.println(\"enter string : \");
String str=s.next();
int i,j;
boolean flag=true;
for( i=0,j=str.length()-1;i<=str.length()/2;i++,j--)
{
if(str.charAt(i)!=str.charAt(j))
flag=false;
}
if(flag==true)
System.out.println(\"palindrome\");
else
System.out.println(\"not a palindrome\");
}
}
Solution
import java.util.Scanner;
public class Palin
{
public static void main(String args[])
{
Scanner s=new Scanner(System.in);
System.out.println(\"enter string : \");
String str=s.next();
int i,j;
boolean flag=true;
for( i=0,j=str.length()-1;i<=str.length()/2;i++,j--)
{
if(str.charAt(i)!=str.charAt(j))
flag=false;
}
if(flag==true)
System.out.println(\"palindrome\");
else
System.out.println(\"not a palindrome\");
}
}.
Factors.javaimport java.io.; import java.util.Scanner; class .pdf
Factors.java
import java.io.*;
import java.util.Scanner;
class Factors
{
public static void main(String args[])
{
int a,i;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter an integer\");
a = in.nextInt();
System.out.println(\"You entered integer \"+a);
System.out.print(\"\ \");
System.out.print(\"The factors are : \");
for(i=1;i<=a/2;i++)
{
if(a%i==0)
System.out.print(i+\",\");
}
System.out.print(a);
}
}
Output:
Enter an integer
44
You entered integer 44
The factors are : 1,2,4,11,22,44
Solution
Factors.java
import java.io.*;
import java.util.Scanner;
class Factors
{
public static void main(String args[])
{
int a,i;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter an integer\");
a = in.nextInt();
System.out.println(\"You entered integer \"+a);
System.out.print(\"\ \");
System.out.print(\"The factors are : \");
for(i=1;i<=a/2;i++)
{
if(a%i==0)
System.out.print(i+\",\");
}
System.out.print(a);
}
}
Output:
Enter an integer
44
You entered integer 44
The factors are : 1,2,4,11,22,44.
Problem1 java codeimport java.util.Scanner; Java code to pr.pdf
// Problem1 java code
import java.util.Scanner;
// Java code to print all possible strings of letter L and R
class Problem1 {
static void print(char set[], int length) {
int setLength = set.length;
printRecursion(set, \"\", setLength, length);
}
static void printRecursion(char set[], String prefixset, int setLength, int length) {
// Base case: length is 0
if (length == 0) {
System.out.println(prefixset);
return;
}
for (int i = 0; i < setLength; ++i) {
String newPrefixset = prefixset + set[i];
printRecursion(set, newPrefixset, setLength, length - 1);
}
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter length: \");
int length = scan.nextInt();
char set[] = {\'L\', \'R\'};
print(set, length);
}
}
/*
output:
Enter length:
3
LLL
LLR
LRL
LRR
RLL
RLR
RRL
RRR
*/
// Problem2 java code
import java.util.Scanner;
// Java code to print all possible strings of letter L and R
class Problem2 {
static void print(char set[], int length) {
int setLength = set.length;
printRecursion(set, \"\", setLength, length);
}
static void printRecursion(char set[], String prefixset, int setLength, int length) {
// Base case: length is 0
if (length == 0) {
System.out.print(prefixset + \" \");
return;
}
for (int i = 0; i < setLength; ++i) {
String newPrefixset = prefixset + set[i];
printRecursion(set, newPrefixset, setLength, length - 1);
}
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter length: \");
int length = scan.nextInt();
char set[] = {\'1\', \'3\', \'5\', \'7\', \'9\'};
print(set, length);
System.out.println();
}
}
/*
output:
Enter length:
3
111 113 115 117 119 131 133 135 137 139 151 153 155
157 159 171 173 175 177 179 191 193 195 197 199 311
313 315 317 319 331 333 335 337 339 351 353 355 357
359 371 373 375 377 379 391 393 395 397 399 511 513
515 517 519 531 533 535 537 539 551 553 555 557 559
571 573 575 577 579 591 593 595 597 599 711 713 715
717 719 731 733 735 737 739 751 753 755 757 759 771
773 775 777 779 791 793 795 797 799 911 913 915 917
919 931 933 935 937 939 951 953 955 957 959 971 973
975 977 979 991 993 995 997 999
*/
// Problem3 java code
import java.util.Scanner;
// Java code to multiply 2 numbers
class Problem3 {
public static int multiply(int a, int b)
{
int temp = b;
for (int i = 1; i < a; i++ ) {
b = b + temp;
}
return b;
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter a: \");
int a = scan.nextInt();
System.out.println(\"Enter b: \");
int b = scan.nextInt();
System.out.println(a + \"X\" + b + \" = \" + multiply(a,b));
}
}
/*
output:
Enter a:
7
Enter b:
6
7X6 = 42
*/
// Problem4 java code
import java.util.Scanner;
// Java code to find gcd 2 numbers
class Problem4 {
public static int gcd(int a, int b)
{
if (b!=0)
return gcd(b, a%b);
else
return a;
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter a: \");
int a = scan.ne.
Nuclear Power Economics and Structuring 2024
Title: Nuclear Power Economics and Structuring - 2024 Edition
Produced by: World Nuclear Association Published: April 2024
Report No. 2024/001
© 2024 World Nuclear Association.
Registered in England and Wales, company number 01215741
This report reflects the views
of industry experts but does not
necessarily represent those
of World Nuclear Association’s
individual member organizations.
Student information management system project report ii.pdf
Our project explains about the student management. This project mainly explains the various actions related to student details. This project shows some ease in adding, editing and deleting the student details. It also provides a less time consuming process for viewing, adding, editing and deleting the marks of the students.
More Related Content
Similar to LAB1.docx
Programs of java
some programs of java language for more info contact 03023688542 Email: shafique.sangi786@gmail.com
CODEimport java.util.; public class test { public static voi.pdf
CODE:
import java.util.*;
public class test {
public static void main(String args[]) {
int num, sum=0;
Scanner input=new Scanner(System.in);
System.out.print(\"Input num : \");
num=input.nextInt();
while(num != 13) {
sum=sum+num;
System.out.print(\"Input num : \");
num=input.nextInt();
}
sum=sum+num;
System.out.println(\"Sum = \"+sum);
}
}
save the above file as test.java
Output:
javac test.java
java test
Input num :
1
Input num :
2
Input num :
3
Input num :
4
Input num :
5
Input num :
6
Input num :
7
Input num :
8
Input num :
9
Input num :
0
Input num :
13
Sum = 58
Solution
CODE:
import java.util.*;
public class test {
public static void main(String args[]) {
int num, sum=0;
Scanner input=new Scanner(System.in);
System.out.print(\"Input num : \");
num=input.nextInt();
while(num != 13) {
sum=sum+num;
System.out.print(\"Input num : \");
num=input.nextInt();
}
sum=sum+num;
System.out.println(\"Sum = \"+sum);
}
}
save the above file as test.java
Output:
javac test.java
java test
Input num :
1
Input num :
2
Input num :
3
Input num :
4
Input num :
5
Input num :
6
Input num :
7
Input num :
8
Input num :
9
Input num :
0
Input num :
13
Sum = 58.
import java.util.;public class Program{public static void.pdf
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
Solution
Hi,
I have modified the code. it is working as expected now. Highlighted the code changes below.
Issue here is with you have written this statement twice System.out.println(x); which causes the
issue.
Program.java
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println();
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Output:
Enter amount
10
Amount is: 10
X
XX
XXX
XXXX
XXXXX
XXXXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
XXXXXXXXXX.
Java programs
Sample java programs for beginners. This slides can help you to print few words in java. Also it helps to do arithmetic calculations and will give you basic concept of for loop, do while loop and if else statements in java. Area of a square.
SumNumbers.java import java.util.Scanner;public class SumNumbe.pdf
SumNumbers.java
import java.util.Scanner;
public class SumNumbers {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
int sum = 0;
while(true){
System.out.println(\"Enter integer:\");
int n = scan.nextInt();
if(n == -1){
break;
}
sum = sum + n;
}
System.out.println(\"The sum is \"+sum);
}
}
Output:
Enter integer:
1
Enter integer:
2
Enter integer:
3
Enter integer:
4
Enter integer:
5
Enter integer:
6
Enter integer:
7
Enter integer:
8
Enter integer:
9
Enter integer:
-1
The sum is 45
Solution
SumNumbers.java
import java.util.Scanner;
public class SumNumbers {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
int sum = 0;
while(true){
System.out.println(\"Enter integer:\");
int n = scan.nextInt();
if(n == -1){
break;
}
sum = sum + n;
}
System.out.println(\"The sum is \"+sum);
}
}
Output:
Enter integer:
1
Enter integer:
2
Enter integer:
3
Enter integer:
4
Enter integer:
5
Enter integer:
6
Enter integer:
7
Enter integer:
8
Enter integer:
9
Enter integer:
-1
The sum is 45.
Codeimport java.util.; class Test{ public static void main(S.pdf
Code:
import java.util.*;
class Test{
public static void main(String args[])throws Exception{
Scanner input = new Scanner(System.in);
System.out.print(\"Enter the First Number : \");
int n1=input.nextInt();
System.out.print(\"Enter the Second Number : \");
int n2=input.nextInt();
System.out.println(\"GCD : \"+gcd(n1,n2));
}
public static int gcd(int n1,int n2){
int r;
while(n2 != 0){
r = n1 % n2;
n1 = n2;
n2 = r;
}
return n1;
}
}
Output:
Enter the First Number : 54
Enter the Second Number : 24
GCD : 6
Solution
Code:
import java.util.*;
class Test{
public static void main(String args[])throws Exception{
Scanner input = new Scanner(System.in);
System.out.print(\"Enter the First Number : \");
int n1=input.nextInt();
System.out.print(\"Enter the Second Number : \");
int n2=input.nextInt();
System.out.println(\"GCD : \"+gcd(n1,n2));
}
public static int gcd(int n1,int n2){
int r;
while(n2 != 0){
r = n1 % n2;
n1 = n2;
n2 = r;
}
return n1;
}
}
Output:
Enter the First Number : 54
Enter the Second Number : 24
GCD : 6.
import java.util.Scanner; public class Palin { public static v.pdf
import java.util.Scanner;
public class Palin
{
public static void main(String args[])
{
Scanner s=new Scanner(System.in);
System.out.println(\"enter string : \");
String str=s.next();
int i,j;
boolean flag=true;
for( i=0,j=str.length()-1;i<=str.length()/2;i++,j--)
{
if(str.charAt(i)!=str.charAt(j))
flag=false;
}
if(flag==true)
System.out.println(\"palindrome\");
else
System.out.println(\"not a palindrome\");
}
}
Solution
import java.util.Scanner;
public class Palin
{
public static void main(String args[])
{
Scanner s=new Scanner(System.in);
System.out.println(\"enter string : \");
String str=s.next();
int i,j;
boolean flag=true;
for( i=0,j=str.length()-1;i<=str.length()/2;i++,j--)
{
if(str.charAt(i)!=str.charAt(j))
flag=false;
}
if(flag==true)
System.out.println(\"palindrome\");
else
System.out.println(\"not a palindrome\");
}
}.
Factors.javaimport java.io.; import java.util.Scanner; class .pdf
Factors.java
import java.io.*;
import java.util.Scanner;
class Factors
{
public static void main(String args[])
{
int a,i;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter an integer\");
a = in.nextInt();
System.out.println(\"You entered integer \"+a);
System.out.print(\"\ \");
System.out.print(\"The factors are : \");
for(i=1;i<=a/2;i++)
{
if(a%i==0)
System.out.print(i+\",\");
}
System.out.print(a);
}
}
Output:
Enter an integer
44
You entered integer 44
The factors are : 1,2,4,11,22,44
Solution
Factors.java
import java.io.*;
import java.util.Scanner;
class Factors
{
public static void main(String args[])
{
int a,i;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter an integer\");
a = in.nextInt();
System.out.println(\"You entered integer \"+a);
System.out.print(\"\ \");
System.out.print(\"The factors are : \");
for(i=1;i<=a/2;i++)
{
if(a%i==0)
System.out.print(i+\",\");
}
System.out.print(a);
}
}
Output:
Enter an integer
44
You entered integer 44
The factors are : 1,2,4,11,22,44.
Problem1 java codeimport java.util.Scanner; Java code to pr.pdf
// Problem1 java code
import java.util.Scanner;
// Java code to print all possible strings of letter L and R
class Problem1 {
static void print(char set[], int length) {
int setLength = set.length;
printRecursion(set, \"\", setLength, length);
}
static void printRecursion(char set[], String prefixset, int setLength, int length) {
// Base case: length is 0
if (length == 0) {
System.out.println(prefixset);
return;
}
for (int i = 0; i < setLength; ++i) {
String newPrefixset = prefixset + set[i];
printRecursion(set, newPrefixset, setLength, length - 1);
}
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter length: \");
int length = scan.nextInt();
char set[] = {\'L\', \'R\'};
print(set, length);
}
}
/*
output:
Enter length:
3
LLL
LLR
LRL
LRR
RLL
RLR
RRL
RRR
*/
// Problem2 java code
import java.util.Scanner;
// Java code to print all possible strings of letter L and R
class Problem2 {
static void print(char set[], int length) {
int setLength = set.length;
printRecursion(set, \"\", setLength, length);
}
static void printRecursion(char set[], String prefixset, int setLength, int length) {
// Base case: length is 0
if (length == 0) {
System.out.print(prefixset + \" \");
return;
}
for (int i = 0; i < setLength; ++i) {
String newPrefixset = prefixset + set[i];
printRecursion(set, newPrefixset, setLength, length - 1);
}
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter length: \");
int length = scan.nextInt();
char set[] = {\'1\', \'3\', \'5\', \'7\', \'9\'};
print(set, length);
System.out.println();
}
}
/*
output:
Enter length:
3
111 113 115 117 119 131 133 135 137 139 151 153 155
157 159 171 173 175 177 179 191 193 195 197 199 311
313 315 317 319 331 333 335 337 339 351 353 355 357
359 371 373 375 377 379 391 393 395 397 399 511 513
515 517 519 531 533 535 537 539 551 553 555 557 559
571 573 575 577 579 591 593 595 597 599 711 713 715
717 719 731 733 735 737 739 751 753 755 757 759 771
773 775 777 779 791 793 795 797 799 911 913 915 917
919 931 933 935 937 939 951 953 955 957 959 971 973
975 977 979 991 993 995 997 999
*/
// Problem3 java code
import java.util.Scanner;
// Java code to multiply 2 numbers
class Problem3 {
public static int multiply(int a, int b)
{
int temp = b;
for (int i = 1; i < a; i++ ) {
b = b + temp;
}
return b;
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter a: \");
int a = scan.nextInt();
System.out.println(\"Enter b: \");
int b = scan.nextInt();
System.out.println(a + \"X\" + b + \" = \" + multiply(a,b));
}
}
/*
output:
Enter a:
7
Enter b:
6
7X6 = 42
*/
// Problem4 java code
import java.util.Scanner;
// Java code to find gcd 2 numbers
class Problem4 {
public static int gcd(int a, int b)
{
if (b!=0)
return gcd(b, a%b);
else
return a;
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter a: \");
int a = scan.ne.
Similar to LAB1.docx (20)
CODEimport java.util.; public class test { public static voi.pdf
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import java.util.;public class Program{public static void.pdf
import java.util.;public class Program{public static void.pdf
SumNumbers.java import java.util.Scanner;public class SumNumbe.pdf
SumNumbers.java import java.util.Scanner;public class SumNumbe.pdf
Codeimport java.util.; class Test{ public static void main(S.pdf
Codeimport java.util.; class Test{ public static void main(S.pdf
WINSEM2020-21_STS3105_SS_VL2020210500169_Reference_Material_I_17-Feb-2021_L8-...
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import java.util.Scanner; public class Palin { public static v.pdf
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Factors.javaimport java.io.; import java.util.Scanner; class .pdf
Factors.javaimport java.io.; import java.util.Scanner; class .pdf
Problem1 java codeimport java.util.Scanner; Java code to pr.pdf
Problem1 java codeimport java.util.Scanner; Java code to pr.pdf
Recently uploaded
Nuclear Power Economics and Structuring 2024
Title: Nuclear Power Economics and Structuring - 2024 Edition
Produced by: World Nuclear Association Published: April 2024
Report No. 2024/001
© 2024 World Nuclear Association.
Registered in England and Wales, company number 01215741
This report reflects the views
of industry experts but does not
necessarily represent those
of World Nuclear Association’s
individual member organizations.
Student information management system project report ii.pdf
Our project explains about the student management. This project mainly explains the various actions related to student details. This project shows some ease in adding, editing and deleting the student details. It also provides a less time consuming process for viewing, adding, editing and deleting the marks of the students.
一比一原版(SFU毕业证)西蒙菲莎大学毕业证成绩单如何办理
SFU毕业证原版定制【微信：176555708】【西蒙菲莎大学毕业证成绩单-学位证】【微信：176555708】（留信学历认证永久存档查询）采用学校原版纸张、特殊工艺完全按照原版一比一制作（包括：隐形水印，阴影底纹，钢印LOGO烫金烫银，LOGO烫金烫银复合重叠，文字图案浮雕，激光镭射，紫外荧光，温感，复印防伪）行业标杆！精益求精，诚心合作，真诚制作！多年品质 ,按需精细制作，24小时接单,全套进口原装设备，十五年致力于帮助留学生解决难题，业务范围有加拿大、英国、澳洲、韩国、美国、新加坡，新西兰等学历材料，包您满意。
◆◆◆◆◆ — — — — — — — — 【留学教育】留学归国服务中心 — — — — — -◆◆◆◆◆
【主营项目】
一.毕业证【微信：176555708】成绩单、使馆认证、教育部认证、雅思托福成绩单、学生卡等！
二.真实使馆公证(即留学回国人员证明,不成功不收费)
三.真实教育部学历学位认证（教育部存档！教育部留服网站永久可查）
四.办理各国各大学文凭(一对一专业服务,可全程监控跟踪进度)
如果您处于以下几种情况：
◇在校期间，因各种原因未能顺利毕业……拿不到官方毕业证【微信：176555708】
◇面对父母的压力，希望尽快拿到；
◇不清楚认证流程以及材料该如何准备；
◇回国时间很长，忘记办理；
◇回国马上就要找工作，办给用人单位看；
◇企事业单位必须要求办理的
◇需要报考公务员、购买免税车、落转户口
◇申请留学生创业基金
留信网认证的作用:
1:该专业认证可证明留学生真实身份
2:同时对留学生所学专业登记给予评定
3:国家专业人才认证中心颁发入库证书
4:这个认证书并且可以归档倒地方
5:凡事获得留信网入网的信息将会逐步更新到个人身份内，将在公安局网内查询个人身份证信息后，同步读取人才网入库信息
6:个人职称评审加20分
7:个人信誉贷款加10分→ 【关于价格问题（保证一手价格）
我们所定的价格是非常合理的，而且我们现在做得单子大多数都是代理和回头客户介绍的所以一般现在有新的单子 我给客户的都是第一手的代理价格，因为我想坦诚对待大家 不想跟大家在价格方面浪费时间
对于老客户或者被老客户介绍过来的朋友，我们都会适当给一些优惠。
8:在国家人才网主办的国家网络招聘大会中纳入资料，供国家高端企业选择人才
选择实体注册公司办理，更放心，更安全！我们的承诺：可来公司面谈，可签订合同，会陪同客户一起到教育部认证窗口递交认证材料，客户在教育部官方认证查询网站查询到认证通过结果后付款，不成功不收费！
学历顾问：微信：176555708
Water Industry Process Automation and Control Monthly - May 2024.pdf
Water Industry Process Automation and Control Monthly - May 2024.pdfWater Industry Process Automation & Control
Welcome to WIPAC Monthly the magazine brought to you by the LinkedIn Group Water Industry Process Automation & Control.
In this month's edition, along with this month's industry news to celebrate the 13 years since the group was created we have articles including
A case study of the used of Advanced Process Control at the Wastewater Treatment works at Lleida in Spain
A look back on an article on smart wastewater networks in order to see how the industry has measured up in the interim around the adoption of Digital Transformation in the Water Industry.Gen AI Study Jams _ For the GDSC Leads in India.pdf
Gen AI Study Jams _ For the GDSC Leads in India.pdf
CFD Simulation of By-pass Flow in a HRSG module by R&R Consult.pptx
CFD analysis is incredibly effective at solving mysteries and improving the performance of complex systems!
Here's a great example: At a large natural gas-fired power plant, where they use waste heat to generate steam and energy, they were puzzled that their boiler wasn't producing as much steam as expected.
R&R and Tetra Engineering Group Inc. were asked to solve the issue with reduced steam production.
An inspection had shown that a significant amount of hot flue gas was bypassing the boiler tubes, where the heat was supposed to be transferred.
R&R Consult conducted a CFD analysis, which revealed that 6.3% of the flue gas was bypassing the boiler tubes without transferring heat. The analysis also showed that the flue gas was instead being directed along the sides of the boiler and between the modules that were supposed to capture the heat. This was the cause of the reduced performance.
Based on our results, Tetra Engineering installed covering plates to reduce the bypass flow. This improved the boiler's performance and increased electricity production.
It is always satisfying when we can help solve complex challenges like this. Do your systems also need a check-up or optimization? Give us a call!
Work done in cooperation with James Malloy and David Moelling from Tetra Engineering.
More examples of our work https://www.r-r-consult.dk/en/cases-en/
J.Yang, ICLR 2024, MLILAB, KAIST AI.pdf
Language-Interfaced Tabular Oversampling Via Progressive Imputation And Self Autentication
Immunizing Image Classifiers Against Localized Adversary Attacks
This paper addresses the vulnerability of deep learning models, particularly convolutional neural networks
(CNN)s, to adversarial attacks and presents a proactive training technique designed to counter them. We
introduce a novel volumization algorithm, which transforms 2D images into 3D volumetric representations.
When combined with 3D convolution and deep curriculum learning optimization (CLO), itsignificantly improves
the immunity of models against localized universal attacks by up to 40%. We evaluate our proposed approach
using contemporary CNN architectures and the modified Canadian Institute for Advanced Research (CIFAR-10
and CIFAR-100) and ImageNet Large Scale Visual Recognition Challenge (ILSVRC12) datasets, showcasing
accuracy improvements over previous techniques. The results indicate that the combination of the volumetric
input and curriculum learning holds significant promise for mitigating adversarial attacks without necessitating
adversary training.
Governing Equations for Fundamental Aerodynamics_Anderson2010.pdf
Governing Equations for Fundamental Aerodynamics
Hierarchical Digital Twin of a Naval Power System
A hierarchical digital twin of a Naval DC power system has been developed and experimentally verified. Similar to other state-of-the-art digital twins, this technology creates a digital replica of the physical system executed in real-time or faster, which can modify hardware controls. However, its advantage stems from distributing computational efforts by utilizing a hierarchical structure composed of lower-level digital twin blocks and a higher-level system digital twin. Each digital twin block is associated with a physical subsystem of the hardware and communicates with a singular system digital twin, which creates a system-level response. By extracting information from each level of the hierarchy, power system controls of the hardware were reconfigured autonomously. This hierarchical digital twin development offers several advantages over other digital twins, particularly in the field of naval power systems. The hierarchical structure allows for greater computational efficiency and scalability while the ability to autonomously reconfigure hardware controls offers increased flexibility and responsiveness. The hierarchical decomposition and models utilized were well aligned with the physical twin, as indicated by the maximum deviations between the developed digital twin hierarchy and the hardware.
The Benefits and Techniques of Trenchless Pipe Repair.pdf
Explore the innovative world of trenchless pipe repair with our comprehensive guide, "The Benefits and Techniques of Trenchless Pipe Repair." This document delves into the modern methods of repairing underground pipes without the need for extensive excavation, highlighting the numerous advantages and the latest techniques used in the industry.
Learn about the cost savings, reduced environmental impact, and minimal disruption associated with trenchless technology. Discover detailed explanations of popular techniques such as pipe bursting, cured-in-place pipe (CIPP) lining, and directional drilling. Understand how these methods can be applied to various types of infrastructure, from residential plumbing to large-scale municipal systems.
Ideal for homeowners, contractors, engineers, and anyone interested in modern plumbing solutions, this guide provides valuable insights into why trenchless pipe repair is becoming the preferred choice for pipe rehabilitation. Stay informed about the latest advancements and best practices in the field.
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Terzaghi's soil bearing capacity theory, developed by Karl Terzaghi, is a fundamental principle in geotechnical engineering used to determine the bearing capacity of shallow foundations. This theory provides a method to calculate the ultimate bearing capacity of soil, which is the maximum load per unit area that the soil can support without undergoing shear failure. The Calculation HTML Code included.
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Water scarcity is the lack of fresh water resources to meet the standard water demand. There are two type of water scarcity. One is physical. The other is economic water scarcity.
一比一原版(IIT毕业证)伊利诺伊理工大学毕业证成绩单专业办理
IIT毕业证原版定制【微信：176555708】【伊利诺伊理工大学毕业证成绩单-学位证】【微信：176555708】（留信学历认证永久存档查询）采用学校原版纸张、特殊工艺完全按照原版一比一制作（包括：隐形水印，阴影底纹，钢印LOGO烫金烫银，LOGO烫金烫银复合重叠，文字图案浮雕，激光镭射，紫外荧光，温感，复印防伪）行业标杆！精益求精，诚心合作，真诚制作！多年品质 ,按需精细制作，24小时接单,全套进口原装设备，十五年致力于帮助留学生解决难题，业务范围有加拿大、英国、澳洲、韩国、美国、新加坡，新西兰等学历材料，包您满意。
◆◆◆◆◆ — — — — — — — — 【留学教育】留学归国服务中心 — — — — — -◆◆◆◆◆
【主营项目】
一.毕业证【微信：176555708】成绩单、使馆认证、教育部认证、雅思托福成绩单、学生卡等！
二.真实使馆公证(即留学回国人员证明,不成功不收费)
三.真实教育部学历学位认证（教育部存档！教育部留服网站永久可查）
四.办理各国各大学文凭(一对一专业服务,可全程监控跟踪进度)
如果您处于以下几种情况：
◇在校期间，因各种原因未能顺利毕业……拿不到官方毕业证【微信：176555708】
◇面对父母的压力，希望尽快拿到；
◇不清楚认证流程以及材料该如何准备；
◇回国时间很长，忘记办理；
◇回国马上就要找工作，办给用人单位看；
◇企事业单位必须要求办理的
◇需要报考公务员、购买免税车、落转户口
◇申请留学生创业基金
留信网认证的作用:
1:该专业认证可证明留学生真实身份
2:同时对留学生所学专业登记给予评定
3:国家专业人才认证中心颁发入库证书
4:这个认证书并且可以归档倒地方
5:凡事获得留信网入网的信息将会逐步更新到个人身份内，将在公安局网内查询个人身份证信息后，同步读取人才网入库信息
6:个人职称评审加20分
7:个人信誉贷款加10分→ 【关于价格问题（保证一手价格）
我们所定的价格是非常合理的，而且我们现在做得单子大多数都是代理和回头客户介绍的所以一般现在有新的单子 我给客户的都是第一手的代理价格，因为我想坦诚对待大家 不想跟大家在价格方面浪费时间
对于老客户或者被老客户介绍过来的朋友，我们都会适当给一些优惠。
8:在国家人才网主办的国家网络招聘大会中纳入资料，供国家高端企业选择人才
选择实体注册公司办理，更放心，更安全！我们的承诺：可来公司面谈，可签订合同，会陪同客户一起到教育部认证窗口递交认证材料，客户在教育部官方认证查询网站查询到认证通过结果后付款，不成功不收费！
学历顾问：微信：176555708
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UofT毕业证原版定制【微信：176555708】【多伦多大学毕业证成绩单-学位证】【微信：176555708】（留信学历认证永久存档查询）采用学校原版纸张、特殊工艺完全按照原版一比一制作（包括：隐形水印，阴影底纹，钢印LOGO烫金烫银，LOGO烫金烫银复合重叠，文字图案浮雕，激光镭射，紫外荧光，温感，复印防伪）行业标杆！精益求精，诚心合作，真诚制作！多年品质 ,按需精细制作，24小时接单,全套进口原装设备，十五年致力于帮助留学生解决难题，业务范围有加拿大、英国、澳洲、韩国、美国、新加坡，新西兰等学历材料，包您满意。
◆◆◆◆◆ — — — — — — — — 【留学教育】留学归国服务中心 — — — — — -◆◆◆◆◆
【主营项目】
一.毕业证【微信：176555708】成绩单、使馆认证、教育部认证、雅思托福成绩单、学生卡等！
二.真实使馆公证(即留学回国人员证明,不成功不收费)
三.真实教育部学历学位认证（教育部存档！教育部留服网站永久可查）
四.办理各国各大学文凭(一对一专业服务,可全程监控跟踪进度)
如果您处于以下几种情况：
◇在校期间，因各种原因未能顺利毕业……拿不到官方毕业证【微信：176555708】
◇面对父母的压力，希望尽快拿到；
◇不清楚认证流程以及材料该如何准备；
◇回国时间很长，忘记办理；
◇回国马上就要找工作，办给用人单位看；
◇企事业单位必须要求办理的
◇需要报考公务员、购买免税车、落转户口
◇申请留学生创业基金
留信网认证的作用:
1:该专业认证可证明留学生真实身份
2:同时对留学生所学专业登记给予评定
3:国家专业人才认证中心颁发入库证书
4:这个认证书并且可以归档倒地方
5:凡事获得留信网入网的信息将会逐步更新到个人身份内，将在公安局网内查询个人身份证信息后，同步读取人才网入库信息
6:个人职称评审加20分
7:个人信誉贷款加10分→ 【关于价格问题（保证一手价格）
我们所定的价格是非常合理的，而且我们现在做得单子大多数都是代理和回头客户介绍的所以一般现在有新的单子 我给客户的都是第一手的代理价格，因为我想坦诚对待大家 不想跟大家在价格方面浪费时间
对于老客户或者被老客户介绍过来的朋友，我们都会适当给一些优惠。
8:在国家人才网主办的国家网络招聘大会中纳入资料，供国家高端企业选择人才
选择实体注册公司办理，更放心，更安全！我们的承诺：可来公司面谈，可签订合同，会陪同客户一起到教育部认证窗口递交认证材料，客户在教育部官方认证查询网站查询到认证通过结果后付款，不成功不收费！
学历顾问：微信：176555708
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- 1. JAVA LAB 1 1: CODE:- import java.util.Scanner; public class Main { public static void main(String[] args) { int n1,n2; Scanner op=new Scanner(System.in); System.out.println("Enter No:"); n1 = op.nextInt(); n2 = op.nextInt(); System.out.println(" Sum of two no:"+(n1+n2)); } }
- 2. OUTPUT:- 3: CODE: import java.util.Scanner; public class Main { public static void main(String[] args) { int l,u; Scanner op=new Scanner(System.in); System.out.println("Enter lower limit and upper limit :"); l = op.nextInt(); u = op.nextInt(); System.out.println("No. in increasing order:"); for(int i=l+1;i<=u-1;i++) { System.out.printf("%d ",i);
- 3. } } } OUTPUT:- CODE 4: import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner op=new Scanner(System.in); int n1,n2,n3,lar=0; System.out.println("Enter three numbers:"); n1 = op.nextInt(); n2 = op.nextInt(); n3 = op.nextInt(); if((n1<=100)&&(n2<=100)&&(n3<=100))
- 4. { lar = (n1>n2)?((n1>n2)?n1:n3):((n2>n3)?n2:n3); System.out.println("Largest of three numbers:"+lar); } else System.out.println("Invalid input"); } } OUTPUT:- CODE 5:- import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner op=new Scanner(System.in);
- 5. int n,temp = 0; System.out.println("Enter number:"); n = op.nextInt(); if(n<=100) { for(int i=n-1;i>=2;i--) { if(n%i==0) { System.out.println("Not Prime"); temp = -1; break; } if(i==2) { if(temp==0) System.out.println("Prime"); else System.out.println("Not Prime");
- 6. } } } else System.out.println(" Invalid input "); } } OUTPUT:- CODE 6:- import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner op=new Scanner(System.in);
- 7. int n1 = 0,n2 = 1,n3=0; System.out.println("Fibonacci series upto 10 terms: "); for(int i=1;i<=10;i++) { n3 = n1+n2; System.out.print(" "+n3); n1=n2; n2=n3; } } } OUTPUT:-