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https://projecteuler.net/problem=4
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
Find the largest palindrome made from the product of two 3-digit numbers.
Here is my code and I find the answer as 580085 but it is not the correct answer:
public class asdas {
public static void main(String[] args) {
int a=100,b=100,answer=0;
while(a<=999)
{
b=100;
while(b<=999)
{
int product=a*b;
int reverse=0;
while(product>0)
{
int lastDigit=product%10;
reverse=(reverse*10)+lastDigit;
product=product/10;
}
product=a*b;
if(product==reverse)
{
answer=product;
}
b++;
}
a++;
}
System.out.println(answer);
}
}
The problem is here:
if(product==reverse)
{
answer=product;
}
You're assuming that if you find a palindrome, it's the largest. That's not true though.
For example, 101*100 is calculated after 100*102, even though the latter is larger. In fact, right after calculating 998*999 (997,002), you calculate 999*100 (99,900), which is nearly 10 times smaller!
You should add another check to make sure the new product is greater:
if(product==reverse && product>answer) {
answer = product;
}
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PROBLEM DESCRIPTION
Peter wants to generate some prime numbers for his cryptosystem. Help him! Your task is to generate all prime numbers between two given numbers!
Input
The input begins with the number t of test cases in a single line (t<=10). In each of the next t lines there are two numbers m and n (1 <= m <= n <= 1000000000, n-m<=100000) separated by a space.
Output
For every test case print all prime numbers p such that m <= p <= n, one number per line, test cases separated by an empty line.
Example
Input:
2
1 10
3 5
Output:
2
3
5
7
//----------BLANK SPACE BETWEEN TEST CASES-------------
3
5
import java.util.*;
import java.lang.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner in = new Scanner(System.in);
int t = in.nextInt();
for(int i=0;i<t;i++)
{
boolean prime = true;
long m = in.nextLong();
long n = in.nextLong();
if(m<2)
{
System.out.println("2");
m=3;
}
for(int j=m;j<=n;j+=2)
{
int z = (int)Math.floor(Math.sqrt(j));
for(int k=2;k<=z;k++)
{
if(j%k==0)
{
prime=false;
break;
}
}
if(prime)
System.out.println(j);
}
System.out.println();
}
}
}
Your loops are using int datatype. It is lossy because you are comparing long with int. It will work for small numbers but constraints are too big so try using long.
Edit
I thought its because overflow without running it.
Your code has lots of problems
It will not work when m is even
you are not setting prime back to true in loop
Check these suggestions and comment is it working?
Solution
use long datatype in loop (that is not necessary)
set prime = true in inner loop
instead of incrementing by 2 i.e. j+=2, increment by 1 j+=1.
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I have this code which I want to use to find prime numbers:
public class Primzahlen {
public static void main(String [] args) {
for(int i = 1;i<100;i++) {
for(int j=1;j<i;j++) {
if(i%j == 0) {break;}
if(j== (i-1)) {System.out.println(i);}
}
}
}
}
But if I try to run the program the output is empty.
The problem with your code is this that, j start from 1 and every number satisfies i%j and the loop breaks so j should start from 2. I would recommend this code which is more effective and runs in O(sqrt(n)) time :
public class Primzahlen {
public static void main(String [] args) {
for(int i = 2;i<100;i++) {
for(int j=2; j < ((int)Math.sqrt(i))+2 ; j++) {
if(i%j == 0) {break;}
if(j== ((int)Math.sqrt(i))+1) {System.out.println(i);}
}
}
}
}
Here are some suggestions.
Seed the first prime as 2. Since 2 is the only even prime this allows subsequent candidates to be just the odd numbers. So starting with 3, increment the candidates by 2.
Only check the primes up to the square root of the candidate.
Since you are finding primes, use the ones already found as the divisors to test for primality. Say the candidate is 17. Primes 2,3,5,7,11,13 have already been found. By applying the previous square root rule, all you need to do is see if 17 is divisible by 3, or 5, and not 2 thru 16. If it were divisible by any prime greater than its square root, it would have already been detected.
For a completely different approach, check out The Sieve of Eratosthenes.
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I tried to write an primenumber generator. The method calcall() should return prime numbers (2,3,5,7...). Unfortunately I get the error, that the method doesn't returns an integer, wich I don't understand. Here is my code:
package primenumber;
public class primecalc {
public static int calcall(int a) { //actual generator
int konstante = a; //is this number a prime num?
int divisor = a-1; //divisor
int var1 = 0; //variable = 0
while(divisor>1) {
int quotient = konstante%divisor; //calc modulo
if(quotient == 0) { //if modulo==0 switch var1 to
var1++; //1 -> no primenumber
break; //stop calculating
} else { //else keep calculculating
divisor--; //until divisor <= 1
}
}
if(var1==0) { //if var1 still 0;
return konstante; //is a primnumber ->
} //return konstante
}
public static void main(String[] args) { //main function
int number = 3; //start with 3
while(True) { //(i'll add 2 manually)
System.out.println(calcall(number)); //print the prime number
number++; //increase number by one
}
}
}
The error is:
Exception in thread "main" java.lang.Error: Unresolved compilation problem:
This method must return a result of type int
at primenumber/primenumber.primecalc.calcall(primecalc.java:5)
at primenumber/primenumber.primecalc.main(primecalc.java:28)
What is wrong?
The gray lines on the code you posted are being ignored by the compiler.
The use of /* and */ makes everything between these seen by the compiler as comments. And that is why those lines are grayed out. If you want to comment on the same line as the code, I'd advise you to use //.
Also, it is common practise to use multi-line comments only to describe functions and place them just above the header of the function. Any other comments should be short, concise and describe functionality. Good variable names and well written code should do most of the explaining, and single line comments should be used when it's a bit harder toperceive what's going on.
Cheers
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I am currently taking a HS course on java, so I am a novice at java to say the least. Right now, for my own use, I am writing a program to take a 2 digit number, and then add up it and all the odd numbers before it till 1. I have the Scanner, calculating whether the number is odd or even, and the runner methods done already(the basic bit), but am a bit confused on the logic. I am trying to use a recursion, and do this code, but am a bit stuck. It would be helpful if you could point me in the right direction, without giving the whole code away. Thanks, - A Novice Programmer
public static void main(String[] args)
{
MathRecursion tester = new MathRecursion();
tester.Method1Runner();
}
public void Method1Runner()
{
GetIntM1();
OddOrEven();
System.out.println("\n\n");
}
public void GetIntM1()
{
Scanner kb = new Scanner(System.in);
System.out.print("\n\n\nEnter a 2 digit integer: ");
twoDig = kb.nextInt();
}
public void OddOrEven()
{
if (twoDig % 2 == 0)
{
//This is even method
Method1a(twoDig);
}
else
{
//This is odd method
Method1b(twoDig);
}
}
public int Method1a(int a)
{
//if (a = 1)
int result = 0;
while (a<=b)
{
result+=a;
a++;
}
System.out.println("The sum of all numbers is "+result);
}
You don't need recursion. The sum of the first n odd numbers is n*n.
The number of odd numbers before a number x is floor(x/2) or in Java (int) x/2 or if x is an int, just x/2.
So the expression in Java that gives you "a 2 digit number, and then add up it and all the odd numbers before it till 1" where the number is stored in int x is:
x + (x/2) * (x/2)
or simplified:
x + x*x/4
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So I have this small method I am working on for class and this question is a tough one and a bit confusing. Basically what I have to to is write a method that will take a number that you input and it will find the next prime number that is 1 greater than a squared number. For example if I were to type in 10 the result would be 17, because 17 is a prime number and its 1 greater than a square, the square being 16. So I have to write a method with only one return statement in it to do this, and I have absolutely no idea where to start. What I have so far ispublic static int nextSquarePlus1Prime(int num){
} Literally. Some helped would be greatly appreciated.
I already have this method where I check if the number is prime or not:
public static boolean isPrime(int num){
int i;
boolean numberIsPrime=true;
if(num==1){
numberIsPrime=false;
}
for(i=2;i*i<=num;i++){
if(num%i==0){
numberIsPrime=false;
}
}
return numberIsPrime;
}
Is there a way to add to this method with another piece of code to work alongside with this one to check if the number is a square?
So this is what i came up with for my code and i put in 10 as my number. I'm getting 50 when I should be getting 17.
public static int nextSquarePlus1Prime(int num){
int i;
int save=0;
for(i=1;i<=num;i++){
if(isPrime(i)&& i*i+1>=num){
save=i*i+1;
}
}
return save;
}
The sqrt() method may help. Here is an idea of how to use it:
int squareRoot = (int) Math.sqrt((double) input);
Input being the number you want to round. Casting the result to an int will automatically round it. It is optional. I cast the input to a double, but you only need to do so if your input in an int.
Here is an easier way to check if a number is prime:
if (input % 2 == 0){
input += 1;
}
return input;
I reccomend you look at #useSticks answer again. Using the approach he described I created a method that does exactly what I think you want.
Here is a method that finds the square root of a positive number:
public static int sqrt(double number){
double g1;
if(number==0){
return 0;
}
double squareRoot = number/2;
do{
g1=squareRoot;
squareRoot = (g1 + (number/g1))/2;
}while((g1-squareRoot)!=0);
return (int) squareRoot;
}
You should provide some additional details for more specific pointers.
In Java, there is a method "is probable prime" that may be of use to you.
Its easier to square a number than it is to find the square root, so it may make sense to find the square root of the starting number, then round up. Take that number and square it, add 1 and check for prime. If not, add 1 to the variable and try again.
Good luck.