import java.util.Scanner;
public class CubesSum {
public static void main (String [] args){
int input;
System.out.println("Enter a positive integer:");
Scanner in = new Scanner(System.in);
input = in.nextInt();
int number = input; //number is a temp variable
int sum = 0;
while(number>0){
int t= number%10;
sum += t*t*t;
number = number/10;
}
System.out.println("The sum of the cubes of the digits is:" +sum);
}
}
Okay so I'm using a while loop. For part B which is to modify to determine what integers of two, three, and four digits are equal to the sum of the cubes of their digits. So for example, 371 = 3³+7³+1³. Can someone tell me how to do it? I need to wrap a for loop around my while loop...
Take the part of your code that computes the sum of the cubes of the digits of a number, and make that a function:
int sumOfCubedDigits(int number) {
int sum = 0;
// compute sum from number
return sum;
}
Then, loop through all the 2-to-4 digit numbers and check whether they equal the sum of the cubes of their digits:
for (int n = 10; n < 10000; n++) {
if (n == sumOfCubedDigits(n)) {
// do whatever with n
}
}
You could keep the sum-of-cubed-digits computation inside the for loop if you want, but it'd be a bit less readable.
Okay, so it looks like you haven't learned about function definitions yet. I shouldn't have assumed. Let's do it with a nested loop, then.
As you said, you need to wrap a for loop around your while. We need to consider all 2-to-4 digit numbers, so our loop will start at the first 2-digit number and end when it reaches the first 5-digit number:
for (int n = 10; n < 10000; n++) {
// More code will go here.
}
Inside the loop, we need to compute the sum of the cubed digits of n. The code you wrote earlier to compute that modifies the number it's operating on, but we can't modify n, or we'll screw up the for loop. We make a copy:
for (int n = 10; n < 10000; n++) {
int temp = n;
int sum = 0;
// Compute the sum of the digits of temp, much like you did before.
}
Finally, if the sum is equal to n, we do something to indicate it. Let's say your assignment said to print all such numbers:
for (int n = 10; n < 10000; n++) {
int temp = n;
int sum = 0;
// Compute the sum of the digits of temp, much like you did before.
if (sum == n) {
System.out.println(n);
}
}
For an arbitrary integer i, it's nth digit dn is, (being n=1 the rightmost digit)
dn = (i % (10^n)) / (10^(n-1)) // all integer operations
as you can see, you'll need to know beforehand the number of digits of your i, otherwise, yes, you'll need a loop
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Please enter number : ");
int num = input.nextInt();
int temp = num, remainder;
int sum = 0;
while(temp %10 != 0){
remainder = temp %10;
sum = sum+ remainder ;
temp = temp/10;
}
System.out.println("Sum of digit : " + sum);
=====OUTPUT====
Please enter number : 123
Sum of digit : 6
Related
I am writing a simple program to find the sum of the odd numbers in between two inputted numbers, and I would appreciate any feedback. I've been working at this problem for a long time, and I could use some expertise.
import java.util.Scanner;
class Main {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
System.out.print("#1: ");
int num1 = s.nextInt();
System.out.print("#2: ");
int num2 = s.nextInt();
int sum = 0;
for(int i= num1; i<(num2-num1); i++){
if (i%2 != 0){
sum+=i;
}
}
System.out.print();
}
}
There are a few simple errors that you can fix in your code here. First, it should not be (num2-num1), but rather just num1. Also, make sure you print your result in the last statement. In contrast, there are many other ways to attack this problem, but if you fix these few issues, your simple method will work just fine.
I would test if num1 is even first, if it is add one to make it odd (otherwise, it's already odd). Then increment by 2, so you know every i is odd. Next, your loop test should be <= num2 (not < num2 - num1), because you want the range from num1 to num2 inclusive. Finally, don't forget to actually print the result. Like,
if (num1 % 2 == 0) {
num1++;
}
int sum = 0;
for (int i = num1; i <= num2; i+=2) {
sum += i;
}
System.out.println(sum);
Alternatively, in Java 8+, you might do it with an IntStream like
System.out.println(IntStream.rangeClosed(num1, num2).filter(x -> x % 2 != 0).sum());
You can also apply some math instead of looping through all the numbers, which is much more efficient and becomes a constant time algorithm instead of linear time.
To get the sum of the numbers you can multiply the average between both numbers by the count of elements in the range.
If any of both numbers is odd, you can add one, or substract one to reach the first even number in the range, so you calculate from an even to an even number.
Example: from 3 to 12, you will use 4 and 12. Average is 8. And you have (12-4)/2 + 1 = 5 even numbers in the range (the +1 is because it's inclusive range). The sum will be 8*5=40.
Which is true: 4+6+8+10+12 = 40
import java.util.Scanner;
class SumOfOdds
{
public static void main(String[] args)
{
Scanner s = new Scanner(System.in);
System.out.print("#1: ");
int num1 = s.nextInt();
System.out.print("#2: ");
int num2 = s.nextInt();
if (num1 % 2 == 0)
{
num1++;
}
int sum = 0;
for (int i = num1; i <= num2; i+=2)
{
sum += i;
}
System.out.println(sum);
}
}
thanks in advance for any help I'm in an intro to java class and our home work was to generate 10 random numbers between 1&50 which I got and then average the generated numbers. I can't figure out how to average them here's what I have. Is there a way to store each random number as a variable?
public class randomNumberGen
{
public static void main(String [] args)
{
Random r=new Random();
for (int i=1;i<=10;i++){
System.out.println(r.nextInt(50));
System.out.println();
int average = (i/4);
System.out.println("your average is"+average);
}
}
}
use streams with java 8
final int numberOfRandom = 10;
final int min = 0;
final int max = 50;
final Random random = new Random();
System.out.println("The ave is: "+random.ints(min, max).limit(numberOfRandom).average());
First of all you have to replace "r.nextInt(50)" for "r.nextInt(50) + 1" because r.nextInt(n) returns a number between 0 (inclusive) and n (exclusive). Then, you know that an average is just a sum of n values divided by n. What you can do is just declare a "total" variable initialized to 0 before the loop. On each iteration you add to this variable the random value generated by r.nextInt(50). After the loop you can just divide the total by 10 so you get the average.
PS: it's a good practice to don't use "magic numbers", so it would be perfect (and luckily your teacher will have it in count) if you declare a constant for the number of iterations and then use it both in the loop condition and in the average calculation. Like this, if you have to make it for 100 numbers you only have to change the constant value from 10 to 100 instead of replacing two 10's por two 100's. Also this gives you the chance to give semantic value to these numbers, because now they will be "AMOUNT_OF_NUMBERS = 10" instead of just "10".
Like every average, it's sum of elements / amount of elements. So let's apply it here:
import java.util.Random;
public class randomNumberGen
{
public static void main(String [] args)
{
Random r=new Random();
double sum = 0; // is double so to prevent int division later on
int amount = 10;
int upperBound = 50;
for (int i = 0; i < amount; i++){
int next = r.nextInt(upperBound) + 1; // creates a random int in [1,50]
System.out.println(next);
sum += next; // accumulate sum of all random numbers
}
System.out.println("Your average is: " + (sum/amount));
}
}
Store variables outside of the loop to store both the total amount of numbers generated as well as the sum of those numbers. After the loop completes, divide the sum by the total amount of numbers.
public static void main(String [] args)
{
Random r=new Random();
double sum = 0;
int totalNums;
for (totalNums=1;totalNums<=10;totalNums++){
int randomNum = r.nextInt(50);
sum += randomNum;
System.out.println(randomNum);
}
double average = sum/totalNums;
System.out.println("your average is: "+average);
}
Average = Sum of numbers / amount of numbers
int sum = 0;
for (int i=1;i<=10;i++){
sum += r.nextInt(50) +1; //nextInt 50 produces value 0 to 49 so you add 1 to get 1 to 50 OR as suggested in the comments sum/10d
}
System.out.println("Average is: " + sum/10) // If you want the result in double (with decimals) just write sum*1.0/10
You could also do the same with a while loop.
int i = 0;
int sum = 0;
while(i < 10){
sum += r.nextInt(50) +1;
i++;
}
System.out.println("Average is: " + sum*1.0/i);
Or even shorter with lambda expressions: (/java 8 streams)
OptionalDouble average = IntStream.range(1, 10).map(x-> x = r.nextInt(50) +1).average();
System.out.println("Average is "+ average.getAsDouble());
.map(x-> x = r.nextInt(50) +1) // maps (changes) each value from 1 to 10 to a random number between 1 and 50
.average(); // calculates the average.
Simply create a variable sum starting at zero that you increment at each iteration. After the loop, simply divide by the number of elements..
Average means you should add everything up and devide it by the number of elements (50).
import java.util.Random;
class Homework {
public static final Random RANDOM = Random(); // never regenerate randoms
public static void main(String args[]) {
final int N = 50;
int sum = 0;
for (int i = 0; i < N; ++i) {
sum += RANDOM.nextInt(50)+1;
}
System.out.println("Avg: "+ sum / (float) N);
}
}
This should do the trick. Try to learn from it not just C+P.
Ps: Friggin annoying to write code on a phone.
I'm kind of a newbie at Java, and not very good at it. It's a trial and error process for me.
I'm working on a Java program to output the amount of primes in an array. I can get it to output the primes, but I want to also output the quantity of primes. I tried to add each prime to an array list titled "primes" then return "primes.size()" at the end of my program. It doesn't work as intended. The count is actually off. When I create an array of 5 numbers, it outputs 3 primes, 2, 3, and 5. But then it says I have 4 primes. I think it might be counting 1 as a prime. Because when I create an array of 20, the prime numbers output 2,3,5,7,11,13,17 and 19. Then it says the total prime numbers = 9. It should be 8 though.
Here's my code
public class Prime {
public static void main(String[] args) {
int index = 0;
Scanner scan = new Scanner(System. in );
System.out.println("How big would you like the array? ");
int num = scan.nextInt();
int[] array = new int[num];
ArrayList < Integer > primes = new ArrayList < Integer > ();
//System.out.println("How Many threads? ");
//int nThreads = scan.nextInt(); // Create variable 'n' to handle whatever integer the user specifies. nextInt() is used for the scanner to expect and Int.
//Thread[] thread = new Thread[nThreads];
for (int n = 1; n <= array.length; n++) {
boolean prime = true;
for (int j = 2; j < n; j++) {
if (n % j == 0) {
prime = false;
break;
}
}
if (prime) {
primes.add(n);
}
if (prime && n != 1) {
System.out.println(n + "");
}
}
System.out.println("Total Prime numbers = " + primes.size());
System.out.println("Prime Numbers within " + array.length);
}
}
Forgive the sloppiness of it. I actually plan on adding multithreading to it, but I wanted to get this down first.
Any help would be greatly appreciated. Thanks.
You have included 1 in your array of primes, because you started the n for loop at 1. You don't print it because of the final if statement, but it's there in the ArrayList.
Start your n for loop with n = 2. As a consequence, you won't need the final if statement, because n won't be 1 ever. You could print the prime at the same time as you add it to the ArrayList.
this is the question, and yes it is homework, so I don't necessarily want anyone to "do it" for me; I just need suggestions: Maximum sum: Design a linear algorithm that finds a contiguous subsequence of at most M in a sequence of N long integers that has the highest sum among all such subsequences. Implement your algorithm, and confirm that the order of growth of its running time is linear.
I think that the best way to design this program would be to use nested for loops, but because the algorithm must be linear, I cannot do that. So, I decided to approach the problem by making separate for loops (instead of nested ones).
However, I'm really not sure where to start. The values will range from -99 to 99 (as per the range of my random number generating program).
This is what I have so far (not much):
public class MaxSum {
public static void main(String[] args){
int M = Integer.parseInt(args[0]);
int N = StdIn.readInt();
long[] a = new long[N];
for (int i = 0; i < N; i++) {
a[i] = StdIn.readLong();}}}
if M were a constant, this wouldn't be so difficult. For example, if M==3:
public class MaxSum2 {
public static void main(String[] args){
int N = StdIn.readInt(); //read size for array
long[] a = new long[N]; //create array of size N
for (int i = 0; i < N; i++) { //go through values of array
a[i] = StdIn.readLong();} //read in values and assign them to
//array indices
long p = a[0] + a[1] + a[2]; //start off with first 3 indices
for (int i =0; i<N-4; i++)
{if ((a[i]+a[i+1]+a[1+2])>=p) {p=(a[i]+a[i+1]+a[1+2]);}}
//if sum of values is greater than p, p becomes that sum
for (int i =0; i<N-4; i++) //prints the subsequence that equals p
{if ((a[i]+a[i+1]+a[1+2])==p) {StdOut.println((a[i]+a[i+1]+a[1+2]));}}}}
If I must, I think MaxSum2 will be acceptable for my lab report (sadly, they don't expect much). However, I'd really like to make a general program, one that takes into consideration the possibility that, say, there could be only one positive value for the array, meaning that adding the others to it would only reduce it's value; Or if M were to equal 5, but the highest sum is a subsequence of the length 3, then I would want it to print that smaller subsequence that has the actual maximum sum.
I also think as a novice programmer, this is something I Should learn to do. Oh and although it will probably be acceptable, I don't think I'm supposed to use stacks or queues because we haven't actually covered that in class yet.
Here is my version, adapted from Petar Minchev's code and with an important addition that allows this program to work for an array of numbers with all negative values.
public class MaxSum4 {
public static void main(String[] args)
{Stopwatch banana = new Stopwatch(); //stopwatch object for runtime data.
long sum = 0;
int currentStart = 0;
long bestSum = 0;
int bestStart = 0;
int bestEnd = 0;
int M = Integer.parseInt(args[0]); // read in highest possible length of
//subsequence from command line argument.
int N = StdIn.readInt(); //read in length of array
long[] a = new long[N];
for (int i = 0; i < N; i++) {//read in values from standard input
a[i] = StdIn.readLong();}//and assign those values to array
long negBuff = a[0];
for (int i = 0; i < N; i++) { //go through values of array to find
//largest sum (bestSum)
sum += a[i]; //and updates values. note bestSum, bestStart,
// and bestEnd updated
if (sum > bestSum) { //only when sum>bestSum
bestSum = sum;
bestStart = currentStart;
bestEnd = i; }
if (sum < 0) { //in case sum<0, skip to next iteration, reseting sum=0
sum = 0; //and update currentStart
currentStart = i + 1;
continue; }
if (i - currentStart + 1 == M) { //checks if sequence length becomes equal
//to M.
do { //updates sum and currentStart
sum -= a[currentStart];
currentStart++;
} while ((sum < 0 || a[currentStart] < 0) && (currentStart <= i));
//if sum or a[currentStart]
} //is less than 0 and currentStart<=i,
} //update sum and currentStart again
if(bestSum==0){ //checks to see if bestSum==0, which is the case if
//all values are negative
for (int i=0;i<N;i++){ //goes through values of array
//to find largest value
if (a[i] >= negBuff) {negBuff=a[i];
bestSum=negBuff; bestStart=i; bestEnd=i;}}}
//updates bestSum, bestStart, and bestEnd
StdOut.print("best subsequence is from
a[" + bestStart + "] to a[" + bestEnd + "]: ");
for (int i = bestStart; i<=bestEnd; i++)
{
StdOut.print(a[i]+ " "); //prints sequence
}
StdOut.println();
StdOut.println(banana.elapsedTime());}}//prints elapsed time
also, did this little trace for Petar's code:
trace for a small array
M=2
array: length 5
index value
0 -2
1 2
2 3
3 10
4 1
for the for-loop central to program:
i = 0 sum = 0 + -2 = -2
sum>bestSum? no
sum<0? yes so sum=0, currentStart = 0(i)+1 = 1,
and continue loop with next value of i
i = 1 sum = 0 + 2 = 2
sum>bestSum? yes so bestSum=2 and bestStart=currentStart=1 and bestEnd=1=1
sum<0? no
1(i)-1(currentStart)+1==M? 1-1+1=1 so no
i = 2 sum = 2+3 = 5
sum>bestSum? yes so bestSum=5, bestStart=currentStart=1, and bestEnd=2
sum<0? no
2(i)-1(currentStart)+1=M? 2-1+1=2 so yes:
sum = sum-a[1(curentstart)] =5-2=3. currentStart++=2.
(sum<0 || a[currentStart]<0)? no
i = 3 sum=3+10=13
sum>bestSum? yes so bestSum=13 and bestStart=currentStart=2 and bestEnd=3
sum<0? no
3(i)-2(currentStart)+1=M? 3-2+1=2 so yes:
sum = sum-a[1(curentstart)] =13-3=10. currentStart++=3.
(sum<0 || a[currentStart]<0)? no
i = 4 sum=10+1=11
sum>bestSum? no
sum<0? no
4(i)-3(currentStart)+1==M? yes but changes to sum and currentStart now are
irrelevent as loop terminates
Thanks again! Just wanted to post a final answer and I was slightly proud for catching the all negative thing.
Each element is looked at most twice (one time in the outer loop, and one time in the while loop).
O(2N) = O(N)
Explanation: each element is added to the current sum. When the sum goes below zero, it is reset to zero. When we hit M length sequence, we try to remove elements from the beginning, until the sum is > 0 and there are no negative elements in the beginning of it.
By the way, when all elements are < 0 inside the array, you should take only the largest negative number. This is a special edge case which I haven't written below.
Beware of bugs in the below code - it only illustrates the idea. I haven't run it.
int sum = 0;
int currentStart = 0;
int bestSum = 0;
int bestStart = 0;
int bestEnd = 0;
for (int i = 0; i < N; i++) {
sum += a[i];
if (sum > bestSum) {
bestSum = sum;
bestStart = currentStart;
bestEnd = i;
}
if (sum < 0) {
sum = 0;
currentStart = i + 1;
continue;
}
//Our sequence length has become equal to M
if (i - currentStart + 1 == M) {
do {
sum -= a[currentStart];
currentStart++;
} while ((sum < 0 || a[currentStart] < 0) && (currentStart <= i));
}
}
I think what you are looking for is discussed in detail here
Find the subsequence with largest sum of elements in an array
I have explained 2 different solutions to resolve this problem with O(N) - linear time.
So the riddle is:
John has written down k sequential odd numbers: n{1}, n{2}, ..., n{k-1}, n{k} (where n{2} = n{1} + 2 and so on). We know that:
The sum of the first four numbers is a fourth power of some prime number (so n{1} + n{2} + n{3} + n{4} = p{1} where p{1}^4 is a prime number.
The sum of the last five numbers is a fourth power of some prime number (so n{k} + n{k-1} + n{k-2} + n{k-3} + n{k-4}= p{2}^4 where p{1} is a prime number.
The question is - how many numbers have been written down (k=?).
Below is my attempt to solve it in Java:
import java.math.BigInteger;
import java.util.Set;
//precalculate prime numbers
public class PrimeSieve {
public static boolean[] calculateIntegers(int N) {
// initially assume all integers are prime
boolean[] isPrime = new boolean[N + 1];
for (int i = 2; i <= N; i++) {
isPrime[i] = true;
}
// mark non-primes <= N using Sieve of Eratosthenes
for (int i = 2; i*i <= N; i++) {
// if i is prime, then mark multiples of i as nonprime
// suffices to consider mutiples i, i+1, ..., N/i
if (isPrime[i]) {
for (int j = i; i*j <= N; j++) {
isPrime[i*j] = false;
}
}
}
return isPrime;
}
}
The solving class:
public class Solver {
static boolean[] isPrime = PrimeSieve.calculateIntegers(100000);
public static void main(String[] args) {
int minNumberCount = 5;
int maxNumberCount = 2000;
int startInt = 2;
int endInt = 1000000;
for (int numberCount = minNumberCount; numberCount < maxNumberCount+1; numberCount++) {
System.out.println("Analyzing for " + numberCount + " numbers");
int[] numbers = new int[numberCount];
//loop through number sets
for (int firstNum = startInt; firstNum < endInt; firstNum+=2) {
//populate numbers array
for(int j=0; j<numberCount; j++){
numbers[j] = firstNum + j*2;
}
long bottomSum=0;
long topSum=0;
//calculate bottom sum
for(int iter=0; iter<4; iter++){
bottomSum+=numbers[iter];
}
//calculate top sum
for(int iter=numberCount-1; iter>numberCount-6; iter--){
topSum+=numbers[iter];
}
//check if the sums match the sulution criteria
if(checkPrime(quadRoot(bottomSum)) && checkPrime(quadRoot(topSum))){
System.out.println("SOLUTION!");
for (int i = 0; i < numbers.length; i++) {
System.out.print(numbers[i] + " ");
}
System.exit(0);
}
}
}
}
private static boolean checkPrime(int i){
return isPrime[i];
}
private static boolean checkPrime(double i){
return ((i % 1) == 0) && checkPrime((int) i);
}
private static double quadRoot(long n){
return Math.sqrt(Math.sqrt(n));
}
}
Using this algorithm with the assumed parameters (max k=2000, max n{1}=100000) - I've found no solution. My question is: are the parameter assumptions wrong (no solution in this range), or do I have some algorithmic/numeric error and that is the reason I've found no solution?
EDIT: sorry - my mistake - it should be ODD instead of EVEN.
It is still easier to solve this directly than to write a program.
The first sum is even so it must be 16 (since 2 is the only even prime). The first four numbers are therefore 1,3,5,7.
The sum of five consecutive odd numbers is 5 times the middle number hence must be divisible by 5. Since it is a fourth power of a prime it must be 625 and the last five numbers are therefore 121,123,125,127,129
It is now an easy task to determine k=65
As said in the comments, your riddle has no solution.
Let's suppose there was a solution, then n1 + n2 + n3 + n4 == p1^4 . We know that n1,n2,n3,n4 are even from the definition of the riddle and therefore as a sum of even numbers, n1 + n2 + n3 + n4 is even as well. This leads us to the fact that p1^4 is even. We know that a multiplication of two odd numbers results only an odd number, hence p1^4 = p1 * p1 * p1 * p1 means that p1 must be an even number. However, p1 is prime. The only prime number which is also even is 2. It's easy to see that there are no four consecutive even numbers that sum up to 16 and therefore p1 is not prime. This contradicts the assumption that p1 is a prime, hence, no solution.
If there are only even numbers, the sum of those is an even number. If I understood correctly, your sum has to be the result of the fourth power of a prime number. Considering the sum is an even number, the only number to satisfy your condition is 16 (2*2*2*2), where 2 is a prime number, so your sum of 4 even number has to be 16. Now, if you're certain there's a sequence, then the sum is computed by adding the first and the last number in the sequence, then multiplying the result with the number of elements in the sequence, and dividing the result of the multiplication by 2. For example, 2+4+6+8=(2+8)*4/2=10*4/2=20. Similarly, for your example, n{1}+n{2}+...+n{k}=(n{1}+n{k})*k/2
On a side note, your smallest sum of 4 even numbers (20), the example I used, is already above your only 4th power of the prime number (16), so yes, there is no valid example in your sequence.
I hope this made some sense