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Project Euler - problem 012

August 1st, 2007 by Daniel Høyer Iversen

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 the 7th triangle number, 28, is the first triangle number to have over five divisors.

Which is the first triangle number to have over five-hundred divisors?

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#include <iostream>
#include <ctime>
using namespace std;
 
#define antDivisors 500
 
bool divisors(int num);
bool isPrimeNumber(int num);
 
int main ()
{
clock_t start = clock();
int res(0), temp(0);
 
while( ! divisors(res) ){
temp++;
res += temp;
}
 
cout << "resultatet er ";
cout << res << "n";
cout<< ( ( clock() - start ) / (double)CLOCKS_PER_SEC ) <<'n';
return 0;
}
 
//from http://mathschallenge.net/index.php?section=faq&ref=number/number_of_divisors
bool divisors(int num) {
int i, ant, factor;
if(num == 0){
return false;
}
i = 2;
factor = 0;
ant = 1;
 
while (num % i == 0) {
factor++;
num /= i;
}
ant *= factor + 1;
factor = 0;
i++;
 
while (i <= num) {
while (!isPrimeNumber(i)) {
i+=2;
}
while (num % i == 0) {
factor++;
num /= i;
}
ant *= factor + 1;
factor = 0;
i+=2;
}
return (ant >= antDivisors);
}
 
bool isPrimeNumber(int num){
if( num % 2 == 0){
return false;
}
 
for(int i = 3; (i-1)*(i-1) < num ; i+=2){
if( num % i == 0){
return false;
}
}
return true;
}
Do you want to use this code?

2 Responses to “Project Euler - problem 012”

  1. schrubi

    57 return (ant >= antDivisors);

    ?? schnall ich net.

    Sep 10th, 2008    11:55 am   
  2. schrubi

    ah, doch. habe die konstantedefinition übersehen

    Sep 10th, 2008    11:56 am   

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