I keep getting an error that the array is out of bounds and I can't figure out what is wrong.
int[] oddArray = new int[500];//holds all the odd numbers
int[] primeArray = new int[500];//holds all the prime numbers
int[] modArray = new int[500];//holds all the mod values
int remainder, p = 0, x = 0;
//fills up the oddArray & modArray
for(int n = 0; n < 500; n++)
{
oddArray[n] = (n * 2) + 1;
modArray[n] = (n* 2) + 1;
}
for(int i = 0; i < 500; i++)
{
//finds prime numbers
for(int n = 0 ; n < 500; n++)
{
//divides the odd numbers by the current mod value
remainder = oddArray[n] % modArray[x];
//if remainder is not 0 it will place a value in prime array
if(remainder != 0)
{
primeArray[p] = oddArray[n];
p++;
}
}
//prints out list of odds/mod/and primes side by side
System.out.println(oddArray[i] + " | " + modArray[i] + " | " + primeArray[p]);
++x;
}
This is the error code
1 | 1 | 0
3 | 3 | 0
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 500
at projectprime_v1.ProjectPrime_V1.main(ProjectPrime_V1.java:41)
Java Result: 1
BUILD SUCCESSFUL (total time: 0 seconds)
You are never resetting p, so eventually it will exceed 499, and cause the exception. Make sure that inside your loop to write p = 0; to reset it. The same goes for x
Hope this helps!
There's this problem from my programming class that I can't get right... The output must be all odd numbers, in odd amounts per line, until the amount of numbers per line meets the odd number that was entered as the input. Example:
input: 5
correct output:
1
3 5 7
9 11 13 15 17
If the number entered is even or negative, then the user should enter a different number. This is what I have so far:
public static void firstNum() {
Scanner kb = new Scanner(System.in);
int num = kb.nextInt();
if (num % 2 == 0 || num < 0)
firstNum();
if (num % 2 == 1)
for (int i = 0; i < num; i++) {
int odd = 1;
String a = "";
for (int j = 1; j <= num; j++) {
a = odd + " ";
odd += 2;
System.out.print(a);
}
System.out.println();
}
}
public static void main(String[] args) {
Scanner kb = new Scanner(System.in);
firstNum();
}
The output I'm getting for input(3) is
1 3 5
1 3 5
1 3 5
When it really should be
1
3 5 7
Can anyone help me?
Try this:
public static void firstNum() {
Scanner kb = new Scanner(System.in);
int num = kb.nextInt();
while (num % 2 == 0 || num < 0) {
num = kb.nextInt();
}
int odd = 1;
for (int i = 1; i <= num; i += 2) {
String a = "";
for (int j = 1; j <= i; j++) {
a = odd + " ";
odd += 2;
System.out.print(a);
}
System.out.println();
}
}
if (num % 2 == 1) {
int odd = 1;
}
for (int i = 0; i < num; i++) {
String a = "";
for (int j = 1; j <= odd; j++) {
a = odd + " ";
odd += 2;
System.out.print(a);
}
System.out.println();
}
You should assign odd before for loop.
In inner for loop compare j and odd together.
For questions like this, usually there is no need to use and conditional statements. Your school probably do not want you to use String as well. You can control everything within a pair of loops.
This is my solution:
int size = 7; // size is taken from user's input
int val = 1;
int row = (size / 2) + 1;
for (int x = 0; x <= row; x++) {
for (int y = 0; y < (x * 2) + 1; y++) {
System.out.print(val + " ");
val += 2;
}
System.out.println("");
}
I left out the part where you need to check whether input is odd.
How I derive my codes:
Observe a pattern in the desired output. It consists of rows and columns. You can easily form the printout by just using 2 loops.
Use the outer loop to control the number of rows. Inner loop to control number of columns to be printed in each row.
The input number is actually the size of the base of your triangle. We can use that to get number of rows.
That gives us: int row = (size/2)+1;
The tricky part is the number of columns to be printed per row.
1st row -> print 1 column
2nd row -> print 3 columns
3rd row -> print 5 columns
4th row -> print 7 columns and so on
We observe that the relation between row and column is actually:
column = (row * 2) + 1
Hence, we have: y<(x*2)+1 as a control for the inner loop.
Only odd number is to be printed, so we start at val 1 and increase val be 2 each time to ensure only odd numbers are printed.
(val += 2;)
Test Run:
1
3 5 7
9 11 13 15 17
19 21 23 25 27 29 31
33 35 37 39 41 43 45 47 49
You can use two nested loops (or streams) as follows: an outer loop through rows with an odd number of elements and an inner loop through the elements of these rows. The internal action is to sequentially print and increase one value.
a loop in a loop
int n = 9;
int val = 1;
// iterate over the rows with an odd
// number of elements: 1, 3, 5...
for (int i = 1; i <= n; i += 2) {
// iterate over the elements of the row
for (int j = 0; j < i; j++) {
// print the current value
System.out.print(val + " ");
// and increase it
val += 2;
}
// new line
System.out.println();
}
a stream in a stream
int n = 9;
AtomicInteger val = new AtomicInteger(1);
// iterate over the rows with an odd
// number of elements: 1, 3, 5...
IntStream.iterate(1, i -> i <= n, i -> i + 2)
// iterate over the elements of the row
.peek(i -> IntStream.range(0, i)
// print the current value and increase it
.forEach(j -> System.out.print(val.getAndAdd(2) + " ")))
// new line
.forEach(i -> System.out.println());
Output:
1
3 5 7
9 11 13 15 17
19 21 23 25 27 29 31
33 35 37 39 41 43 45 47 49
See also: How do I create a matrix with user-defined dimensions and populate it with increasing values?
Seems I am bit late to post, here is my solution:
public static void firstNum() {
Scanner scanner = new Scanner(System.in);
System.out.println("Enter the odd number: ");
int num = scanner.nextInt();
if (num % 2 == 0 || num < 0) {
firstNum();
}
if (num % 2 == 1) {
int disNum = 1;
for (int i = 1; i <= num; i += 2) {
for (int k = 1; k <= i; k++, disNum += 2) {
System.out.print(disNum + " ");
}
System.out.println();
}
}
}
This is the question:
Design and implement an application that creates a histogram that allows you to
visually inspect the frequency distribution of a set of values. The program should read in
an arbitrary number of integers that are in the range 1 to 100 inclusive; then produce a
chart similar to the one below that indicates how many input values fell in the range 1
to 10, 11 to 20, and so on. Print one asterisk for each value entered.
1 - 10 | *****
11 - 20 | **
21 - 30 | *******************
31 - 40 |
41 - 50 | ***
51 - 60 | ********
61 - 70 | **
71 - 80 | *****
81 - 90 | *******
91 - 100 | *********
Source Code:
import java.io.*;
import java.util.Scanner;
public class histogram
{
public static void main(java.lang.String[] args)
throws IOException
{
Scanner scan = new Scanner (System.in);
final int min = 1;
final int max = 10;
final int limit = 10;
int[] a = new int[max];
for (int b = 0; b < a.length; b++)
{
a[b] = 0;
}
System.out.println("Enter Number between 1 and 100: \n (Or press 0 to stop)");
int number = scan.nextInt();
while (number >= min && number <= (limit*max) && number != 0)
{
a[(number-1)/limit] = a[(number - 1 ) / limit] + 1;
System.out.print("Please enter a value:");
number = scan.nextInt();
}
System.out.println("\n__Histogram__");
for (int y = 0; y < a.length; y++)
{
System.out.print(" " + (y * limit + 1) + " - " + (y + 1) * limit + "\t");
return;
}
for (int z = 0; z < a[b]; z++)
{
System.out.print("*");
}
System.out.println(0);
}
}
In the last for statement, it says that b cannot be resolved to a variable. When I use java for help it sets b to 0 and the program doesn't run correctly. Any ideas?
Since b was an array index in the earlier loop, I think you will need to put the last loop inside the loop to increment through the array indexes as you did in the beginning.
Changed b to y and moved z for loop inside y for loop.
for (int y = 0; y < a.length; y++)
{
System.out.print(" " + (y * limit + 1) + " - " + (y + 1) * limit + "\t");
for (int z = 0; z < a[y]; z++)
{
System.out.print("*");
}
System.out.println();
}
I wanted to count sum and length of the longest subsequence in the given array t.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class so {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] s = br.readLine().split(" ");
br.close();
int[] t = new int[s.length];
for (int i = 0; i < s.length; i++) {
t[i] = Integer.parseInt(s[i]);
}
int length = 1;
int sum = t[0];
int maxSum = 0;
int maxLength = 0;
for (int i = 1; i < t.length; i++) {
for (; i < t.length && t[i - 1] <= t[i]; i++) {
length++;
sum += t[i];
System.out.print(t[i] + " ");
}
if (length > maxLength) {
maxLength = length;
maxSum = sum;
length = 1;
sum = 0;
i--;
}
}
System.out.println("sum is " + maxSum + " length is " + maxLength);
}
}
for the numbers1 1 7 3 2 0 0 4 5 5 6 2 1 I get output:
sum is 20 length is 6
but for the same numbers in reverse order 1 2 6 5 5 4 0 0 2 3 7 1 1 I get output:
sum is 17 length is 6 which is not true because I should get sum is 12 length is 5.
Could someone spot the my mistake?
You are reseting the length and sum only when you find the next longest sequence but you should reset them every time you finish testing a sequence:
Right now, your code accumulates length and sum until it surpasses maxLength but length and sum are test variables that need to be reset when testing each possible subsequence.
Furthermore, you sum variable needs to reset to the current test value at t[i - 1] and not to 0. The reason you are getting a correct result even though this bug is present is because the first item in the LIS for both of your inputs is 0.
If we input something like (replace two 0s in the first input with 1s):
1 1 7 3 2 1 1 4 5 5 6 2 1
Output is:
sum is 21 length is 6
But sum should be 22
In fact, a slightly cleaner way would be to perform initialization of your test variables at the beginning of the loop instead of initializing outside of the loop and then reseting inside:
// ...
int length, sum, maxSum = Integer.MIN_VALUE, maxLength = Integer.MIN_VALUE;
for (int i = 1; i < t.length; i++) {
// initialize test variables
length = 1;
sum = t[i - 1];
for (; i < t.length && t[i - 1] <= t[i]; i++) {
length++;
sum += t[i];
System.out.print(t[i] + " ");
}
if (length > maxLength) {
maxLength = length;
maxSum = sum;
i--;
}
}
// ...
NOTE: I added initialization for maxLength and maxSum to use the smallest possible integer to allow counting negative numbers as well.
Ok guys, so I'm doing the Project Euler challenges and I can't believe I'm stuck on the first challenge. I really can't see why I'm getting the wrong answer despite my code looking functional:
import java.util.ArrayList;
public class Multithree {
public static void main(String[] args) {
// TODO Auto-generated method stub
ArrayList<Integer> x = new ArrayList<Integer>();
ArrayList<Integer> y = new ArrayList<Integer>();
int totalforthree = 0;
int totalforfive = 0;
int total =0;
for(int temp =0; temp < 1000 ; temp++){
if(temp % 3 == 0){
x.add(temp);
totalforthree += temp;
}
}
for(int temp =0; temp < 1000 ; temp++){
if(temp % 5 == 0){
y.add(temp);
totalforfive += temp;
}
}
total = totalforfive + totalforthree;
System.out.println("The multiples of 3 or 5 up to 1000 are: " +total);
}
}
I'm getting the answer as 266333 and it says it's wrong...
you should use the same for loop for both to aviod double counting numbers that are multiple of both. such as 15,30...
for(int temp =0; temp < 1000 ; temp++){
if(temp % 3 == 0){
x.add(temp);
totalforthree += temp;
}else if(temp % 5 == 0){
y.add(temp);
totalforfive += temp;
}
}
In a mathematical perspective,
You did not consider about common factors between 3 and 5.Because there is double counting.
ex; number 15 ,
30 ,
45 ,
60 ,
75 ,
90 ,
105 ,
120 ,
135 ,
150 ,
165 ,
180 ,
195 ,
210 ,
225 ,
240 ,
255 ,
270 ,
285 ,
300 ,
315 ,
330 ,
345 ,
360 ,
375 ,
390 ,
405 ,
420 ,
435 ,
450 ,
465 ,
480 ,
495 ,
510 ,
525 ,
540 ,
555 ,
570 ,
585 ,
600 ,
615 ,
630 ,
645 ,
660 ,
675 ,
690 ,
705 ,
720 ,
735 ,
750 ,
765 ,
780 ,
795 ,
810 ,
825 ,
840 ,
855 ,
870 ,
885 ,
900 ,
915 ,
930 ,
945 ,
960 ,
975 ,
990 , are common factors.
total of common factors = 33165.
Your answer is 266333
Correct answer is 233168.
Your Answer - Total of common factors
266333-33165=233168.
(this is a code for getting common factors and Total of common factors )
public static void main(String[] args) {
System.out.println("The sum of the Common Factors : " + getCommonFactorSum());
}
private static int getCommonFactorSum() {
int sum = 0;
for (int i = 1; i < 1000; i++) {
if (i % 3 == 0 && i % 5 == 0) {
sum += i;
System.out.println(i);
}
}
Don't you all think instead of using loops to compute the sum of multiples, we can easily compute sum of n terms using a simple formula of Arithmetic Progression to compute sum of n terms.
I evaluated results on both using loop and formula. Loops works simply valuable to short data ranges. But when the data ranges grows more than 1010 program takes more than hours to process the result with loops. But the same evaluates the result in milliseconds when using a simple formula of Arithmetic Progression.
What we really need to do is:
Algorithm :
Compute the sum of multiples of 3 and add to sum.
Compute the sum of multiples of 5 and add to sum.
Compute the sum of multiples of 3*5 = 15 and subtract from sum.
Here is code snippet in java from my blog post CodeForWin - Project Euler 1: Multiples of 3 and 5
n--; //Since we need to compute the sum less than n.
//Check if n is more than or equal to 3 then compute sum of all divisible by
//3 and add to sum.
if(n>=3) {
totalElements = n/3;
sum += (totalElements * ( 3 + totalElements*3)) / 2;
}
//Check if n is more than or equal to 5 then compute sum of all elements
//divisible by 5 and add to sum.
if(n >= 5) {
totalElements = n/5;
sum += (totalElements * (5 + totalElements * 5)) / 2;
}
//Check if n is more than or equal to 15 then compute sum of all elements
//divisible by 15 and subtract from sum.
if(n >= 15) {
totalElements = n/15;
sum -= (totalElements * (15 + totalElements * 15)) / 2;
}
System.out.println(sum);
If you are using Java 8 you can do it in the following way:
Integer sum = IntStream.range(1, 1000) // create range
.filter(i -> i % 3 == 0 || i % 5 == 0) // filter out
.sum(); // output: 233168
To count the numbers which are divisible by both 3 and 5 twice you can
either write the above line twice or .map() the 2 * i values:
Integer sum = IntStream.range(1, 1000)
.filter(i -> i % 3 == 0 || i % 5 == 0)
.map(i -> i % 3 == 0 && i % 5 == 0 ? 2 * i : i)
.sum(); // output: 266333
How I solved this is that I took an integer value (initialized to zero) and kept on adding the incremented value of i, if its modulo with 3 or 5 gives me zero.
private static int getSum() {
int sum = 0;
for (int i = 1; i < 1000; i++) {
if (i % 3 == 0 || i % 5 == 0) {
sum += i;
}
}
return sum;
}
I did this several ways and several times. The fastest, cleanest and simplest way to complete the required code for Java is this:
public class MultiplesOf3And5 {
public static void main(String[] args){
System.out.println("The sum of the multiples of 3 and 5 is: " + getSum());
}
private static int getSum() {
int sum = 0;
for (int i = 1; i < 1000; i++) {
if (i % 3 == 0 || i % 5 == 0) {
sum += i;
}
}
return sum;
}
If anyone has a suggestion to get it down to fewer lines of code, please let me know your solution. I'm new to programming.
int count = 0;
for (int i = 1; i <= 1000 / 3; i++)
{
count = count + (i * 3);
if (i < 1000 / 5 && !(i % 3 == 0))
{
count = count + (i * 5);
}
}
Others have already pointed out the mistakes in your code, however I want to add that the modulus operator solution is not the most efficient.
A faster implementation would be something like this:
int multiply3_5(int max)
{
int i, x3 = 0, x5 = 0, x15 = 0;
for(i = 3; i < max; i+=3) x3 += i; // Store all multiples of 3
for(i = 5; i < max; i+=5) x5 += i; // Store all multiples of 5
for(i = 15; i < max; i+=15) x15 += i; // Store all multiples 15;
return x3+x5-x15;
}
In this solution I had to take out multiples of 15 because, 3 and 5 have 15 as multiple so on the second loop it will add multiples of 15 that already been added in the first loop;
Solution with a time complexity of O(1)
Another even better solution (with a time complexity of O(1)) is if you take a mathematical approach.
You are trying to sum all numbers like this 3 + 6 + 9 ... 1000 and 5 + 10 + 15 +20 + ... 1000 this is the same of having 3 * (1 + 2 + 3 + 4 + … + 333) and 5 * ( 1 + 2 + 3 + 4 + ... + 200), the sum of 'n' natural number is (n * (n + 1)) (source) so you can calculate in a constant time, as it follows:
int multiply3_5(int max)
{
int x3 = (max - 1) / 3;
int x5 = (max - 1) / 5;
int x15 = (max - 1) / 15;
int sn3 = (x3 * (x3 + 1)) / 2;
int sn5 = (x5 * (x5 + 1)) / 2;
int sn15 = (x15 * (x15 + 1)) / 2;
return (3*sn3) + (5 *sn5) - (15*sn15);
}
Running Example:
public class SumMultiples2And5 {
public static int multiply3_5_complexityN(int max){
int i, x3 = 0, x5 = 0, x15 = 0;
for(i = 3; i < max; i+=3) x3 += i; // Store all multiples of 3
for(i = 5; i < max; i+=5) x5 += i; // Store all multiples of 5
for(i = 15; i < max; i+=15) x15 += i; // Store all multiples 15;
return x3 + x5 - x15;
}
public static int multiply3_5_constant(int max){
int x3 = (max - 1) / 3;
int x5 = (max - 1) / 5;
int x15 = (max - 1) / 15;
int sn3 = (x3 * (x3 + 1)) / 2;
int sn5 = (x5 * (x5 + 1)) / 2;
int sn15 = (x15 * (x15 + 1)) / 2;
return (3*sn3) + (5 *sn5) - (15*sn15);
}
public static void main(String[] args) {
System.out.println(multiply3_5_complexityN(1000));
System.out.println(multiply3_5_constant(1000));
}
}
Output:
233168
233168
Just simply
public class Main
{
public static void main(String[] args) {
int sum=0;
for(int i=1;i<1000;i++){
if(i%3==0 || i%5==0){
sum+=i;
}
}
System.out.println(sum);
}
}
You are counting some numbers twice. What you have to do is add inside one for loop, and use an if-else statement where if you find multiples of 3, you do not count them in 5 as well.
if(temp % 3 == 0){
x.add(temp);
totalforthree += temp;
} else if(temp % 5 == 0){
y.add(temp);
totalforfive += temp;
}
Logics given above are showing wrong answer, because multiples of 3 & 5 are taken for calculation. There is something being missed in above logic, i.e., 15, 30, 45, 60... are the multiple of 3 and also multiple of 5. then we need to ignore such while adding.
public static void main(String[] args) {
int Sum=0, i=0, j=0;
for(i=0;i<=1000;i++)
if (i%3==0 && i<=999)
Sum=Sum+i;
for(j=0;j<=1000;j++)
if (j%5==0 && j<1000 && j*5%3!=0)
Sum=Sum+j;
System.out.println("The Sum is "+Sum);
}
Okay, so this isn't the best looking code, but it get's the job done.
public class Multiples {
public static void main(String[]args) {
int firstNumber = 3;
int secondNumber = 5;
ArrayList<Integer> numberToCheck = new ArrayList<Integer>();
ArrayList<Integer> multiples = new ArrayList<Integer>();
int sumOfMultiples = 0;
for (int i = 0; i < 1000; i++) {
numberToCheck.add(i);
if (numberToCheck.get(i) % firstNumber == 0 || numberToCheck.get(i) % secondNumber == 0) {
multiples.add(numberToCheck.get(i));
}
}
for (int i=0; i<multiples.size(); i++) {
sumOfMultiples += multiples.get(i);
}
System.out.println(multiples);
System.out.println("Sum Of Multiples: " + sumOfMultiples);
}
}
public class Solution {
public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t>0){
int sum = 0;
int count =0;
int n = sc.nextInt();
n--;
System.out.println((n/3*(6+(n/3-1)*3))/2 + (n/5*(10+(n/5-1)*5))/2 - (n/15*(30+(n/15-1)*15))/2);
t--;
}
}
}
If number is 10 then multiple of 3 is 3,6,9 and multiple of 5 is 5,10 total sum is 33 and program gives same answer:
package com.parag;
/*
* #author Parag Satav
*/
public class MultipleAddition {
/**
* #param args
*/
public static void main( final String[] args ) {
// TODO Auto-generated method stub
ArrayList<Integer> x = new ArrayList<Integer>();
ArrayList<Integer> y = new ArrayList<Integer>();
int totalforthree = 0;
int totalforfive = 0;
int number = 8;
int total = 0;
for ( int temp = 1; temp <= number; temp++ ) {
if ( temp % 3 == 0 ) {
x.add( temp );
totalforthree += temp;
}
else if ( temp % 5 == 0 ) {
y.add( temp );
totalforfive += temp;
}
}
total = totalforfive + totalforthree;
System.out.println( "multiples of 3 : " + x );
System.out.println( "multiples of 5 : " + y );
System.out.println( "The multiples of 3 or 5 up to " + number + " are: " + total );
}
}