I am trying to solve Problem 12 on Project Euler. I have an idea of how to complete this problem, however I encountered an error. I've searched how to use arraylist from different questions however I am still facing problems.
import java.util.ArrayList;
public class Level_12 {
/*
The sequence of triangle numbers is generated by adding the natural numbers.
So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
*/
public static ArrayList<Long> check(long num) {
ArrayList<Long> divisors = new ArrayList<Long>();
for (long o = 1; o <= Math.sqrt(num); o++)
if (num % o == 0) {
divisors.add(o);
System.out.println(o + " is a current divisor of " + num);
}
for (Long m : divisors) {
return m;
}
}
public static void main(String[] args) {
long triangle = 0; //Triangle number
long total = 500; //Total divisors
long currenttotal = 0; //Amount of divisors
long i = 0; //Just to itterate
while (currenttotal <= total) {
if (check(i) > currenttotal) { //Finding if the
triangle = i;
if (currenttotal == total) break;
}
i++;
}
System.out.println("The value of the first triangle number to have over 500 divisors is " + triangle);
}
}
EDIT:
Fixed with final code as
import java.util.ArrayList;
public class Level_12 {
/*
The sequence of triangle numbers is generated by adding the natural numbers.
So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
*/
public static ArrayList<Long> check(long num) {
ArrayList<Long> divisors = new ArrayList<Long>();
for (long o = 1; o <= Math.sqrt(num); o++)
if (num % o == 0) {
divisors.add(o);
System.out.println(o + " is a current divisor of " + num);
}
return divisors;
}
public static void main(String[] args) {
long triangle = 0; //Triangle number
long total = 500; //Total divisors
long currenttotal = 0; //Amount of divisors
long i = 0; //Just to itterate
while (currenttotal <= total) {
if (check(i).size() > currenttotal) { //Finding if the amount of divisors is larger than current divisors
triangle = i;
if (currenttotal == total) break;
}
i++;
}
System.out.println("The value of the first triangle number to have over 500 divisors is " + triangle);
}
}
您在check()
方法中返回Long
而不是ArrayList<Long>
Returning Long
while the return type of method is ArrayList<Long>
.So update the return type as below :
public static Long check(long num) {
ArrayList<Long> divisors = new ArrayList<Long>();
for (long o = 1; o <= Math.sqrt(num); o++)
if (num % o == 0) {
divisors.add(o);
System.out.println(o + " is a current divisor of " + num);
}
for (Long m : divisors) {
return m;
}
}
The culprit is this :
for (Long m : divisors) {
return m;
}
In this you are returning a Long value whereas the method signature specifies returning ArrayList
public static **ArrayList<Long>** check(long num)
You need to either change the method signature or return the appropriate value depending on what you want to accomplish.
You are only putting half the divisors of m
into the list. If o
divides m
, then m/o
divides m
too.
/**
* Return a list of the divisors of an integer. If it is a perfect square, the square
* root will be repeated. The list will not be in order.
* @param num The number whose divisors to find.
* @return A list of the number's positive divisors.
*/
public List<Long> divisors(long num) {
final List<Long> divisors = new ArrayList<>();
final long limit = Math.sqrt(num);
for (long o = 1L; o <= limit; o++)
if (num % o == 0) {
divisors.add(o);
divisors.add(num / o);
}
return divisors;
}
/**
* Return the nth triangular number.
* @param n The index of the desired triangular number.
* @return The n-th triangular number 1 + 2 + ... + n
*/
public long triangle(long n) {
return n * (n + 1) / 2;
}
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