Finding a squence.May 12, 2007Problem #14 at project euler is straightforward. You need to find an iterative sequence or rather the starting number that results in the most terms. If you are a strict mathematician the problem description isn't very sound but anyway here is the solution:
public void problem14()
{
int n0;
long n;
int maxTerms=0, terms=0, nMax=0;
int results[] = new int[1000002];
for(n0=5 ; n0 < 1000000 ; n0++, terms=0)
{
n = n0;
do {
if(n % 2 == 0)
{
n = n/2;
}
else
{
n = 3*n + 1;
}
terms++;
//System.out.print(n +",");
if(n < n0)
{
if(n < 0)
{
System.out.println("N =" + n +" for n0="+n0);
break;
}
if(results[(int)n] != 0)
{
terms += results[(int)n];
break;
}
}
}while(n != 1);
results[n0] = terms;
if(maxTerms < terms)
{
maxTerms = terms;
nMax = n0;
}
//System.out.println("");
}
System.out.println(nMax +" gives the most terms" + maxTerms);
}
At first I used integers instead of longss and ran into a situation where some of the terms exceeded 2^32 -1 which lead to array out of bounds exceptions. Switching to longs solved that one easily. Posted by raditha at May 12, 2007 10:42 AM
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