I am trying to complete Project Euler Problem 14 in c++ and I am honestly stuck. Right now when I run the problem it gets stuck at So Far: the number with the highest count: 113370 with the count of 155
So Far: the number with the highest count but when I try changing the i value to over 113371 it works. What is going on??
The question is:
The following iterative sequence is defined for the set of positive
integers: n → n/2 (n is even) n → 3n + 1 (n is odd)
Using the rule above and starting with 13, we generate the following
sequence:
13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1 It can be seen that this
sequence (starting at 13 and finishing at 1) contains 10 terms.
Although it has not been proved yet (Collatz Problem), it is
thought that all starting numbers finish at 1. Which starting number,
under one million, produces the longest chain?
#include<stdio.h>
int main() {
int limit = 1000000;
int highNum, number, i;
int highCount = 0;
int count = 0;
for( number = 13; number <= 1000000; number++ )
{
i = number;
while( i != 1 ) {
if (( i % 2 ) != 0 ) {
i = ( i * 3 ) + 1;
count++;
}
else {
count++;
i /= 2;
}
}
count++;
printf( "So Far: the number with the highest count: %d with the count of %d\n",
number, count );
if( highCount < count ) {
highCount = count;
highNum = number;
}
count = 0;
//break;
}
printf( "The number with the highest count: %d with the count of %d\n",
highNum, highCount );
}
You are getting integer overflow. Update your code like this and see it yourself:
if (( i % 2 ) != 0 ) {
int prevI = i;
i = ( i * 3 ) + 1;
if (i < prevI) {
printf("oops, i < prevI: %d\n", i);
return 0;
}
count++;
}
You should change the type of i to long long or unsigned long long to prevent the overflow.
(And yes, cache the intermediate results)
Remember all intermediate results (up to some suitably high number).
Also, use a big-enough type:
#include <stdio.h>
static int collatz[4000000];
unsigned long long collatzmax;
int comp(unsigned long long i) {
if(i>=sizeof collatz/sizeof*collatz) {
if(i>collatzmax)
collatzmax = i;
return 1 + comp(i&1 ? 3*i+1 : i/2);
}
if(!collatz[i])
collatz[i] = 1 + comp(i&1 ? 3*i+1 : i/2);
return collatz[i];
}
int main() {
collatz[1] = 1;
int highNumber= 1, highCount = 1, c;
for(int i = 2; i < 1000000; i++)
if((c = comp(i)) > highCount) {
highCount = c;
highNumber = i;
}
printf( "The number with the highest count: %d with the count of %d\n",
highNumber, highCount );
printf( "Highest intermediary number: %llu\n", collatzmax);
}
On coliru: http://coliru.stacked-crooked.com/a/773bd8c5f4e7d5a9
Variant with smaller runtime: http://coliru.stacked-crooked.com/a/2132cb74e4605d5f
The number with the highest count: 837799 with the count of 525
Highest intermediary number: 56991483520
BTW: The highest intermediary encountered needs 36 bit to represent as an unsigned number.
With your algorithm, you compute several time identical series.
you may cache result for previous numbers and reuse them.
Something like:
void compute(std::map<std::uint64_t, int>& counts, std::uint64_t i)
{
std::vector<std::uint64_t> series;
while (counts[i] == 0) {
series.push_back(i);
if ((i % 2) != 0) {
i = (i * 3) + 1;
} else {
i /= 2;
}
}
int count = counts[i];
for (auto it = series.rbegin(); it != series.rend(); ++it)
{
counts[*it] = ++count;
}
}
int main()
{
const std::uint64_t limit = 1000000;
std::map<std::uint64_t, int> counts;
counts[1] = 1;
for (std::size_t i = 2; i != limit; ++i) {
compute(counts, i);
}
auto it = std::max_element(counts.begin(), counts.end(),
[](const std::pair<std::uint64_t, int>& lhs, const std::pair<std::uint64_t, int>& rhs)
{
return lhs.second < rhs.second;
});
std::cout << it->first << ":" << it->second << std::endl;
std::cout << limit-1 << ":" << counts[limit-1] << std::endl;
}
Demo (10 seconds)
Don't recompute the same intermediate results over and over!
Given
typedef std::uint64_t num; // largest reliable built-in unsigned integer type
num collatz(num x)
{
return (x & 1) ? (3*x + 1) : (x/2);
}
Then the value of collatz(x) only depends on x, not on when you call it. (In other words, collatz is a pure function.) As a consequence, you can memoize the values of collatz(x) for different values of x. For this purpose, you could use a std::map<num, num> or a std::unordered_map<num, num>.
For reference, here is the complete solution.
And here it is on Coliru, with timing (2.6 secs).
Related
So I was inspired by a recent Youtube video from the Numberphile Channel. This one to be exact. Cut to around the 5 minute mark for the exact question or example that I am referring to.
TLDR; A number is created with all the digits corresponding to 1 to N. Example: 1 to 10 is the number 12,345,678,910. Find out if this number is prime. According to the video, N has been checked up to 1,000,000.
From the code below, I have taken the liberty of starting this process at 1,000,000 and only going to 10,000,000. I'm hoping to increase this to a larger number later.
So my question or the assistance that I need is optimization for this problem. I'm sure each number will still take very long to check but even a minimal percentage of optimization would go a long way.
Edit 1: Optimize which division numbers are used. Ideally this divisionNumber would only be prime numbers.
Here is the code:
#include <iostream>
#include <chrono>
#include <ctime>
namespace
{
int myPow(int x, int p)
{
if (p == 0) return 1;
if (p == 1) return x;
if (p == 2) return x * x;
int tmp = myPow(x, p / 2);
if (p % 2 == 0) return tmp * tmp;
else return x * tmp * tmp;
}
int getNumDigits(unsigned int num)
{
int count = 0;
while (num != 0)
{
num /= 10;
++count;
}
return count;
}
unsigned int getDigit(unsigned int num, int position)
{
int digit = num % myPow(10, getNumDigits(num) - (position - 1));
return digit / myPow(10, getNumDigits(num) - position);
}
unsigned int getTotalDigits(int num)
{
unsigned int total = 0;
for (int i = 1; i <= num; i++)
total += getNumDigits(i);
return total;
}
// Returns the 'index'th digit of number created from 1 to num
int getIndexDigit(int num, int index)
{
if (index <= 9)
return index;
for (int i = 10; i <= num; i++)
{
if (getTotalDigits(i) >= index)
return getDigit(i, getNumDigits(i) - (getTotalDigits(i) - index));
}
}
// Can this be optimized?
int floorSqrt(int x)
{
if (x == 0 || x == 1)
return x;
int i = 1, result = 1;
while (result <= x)
{
i++;
result = i * i;
}
return i - 1;
}
void PrintTime(double num, int i)
{
constexpr double SECONDS_IN_HOUR = 3600;
constexpr double SECONDS_IN_MINUTE = 60;
double totalSeconds = num;
int hours = totalSeconds / SECONDS_IN_HOUR;
int minutes = (totalSeconds - (hours * SECONDS_IN_HOUR)) / SECONDS_IN_MINUTE;
int seconds = totalSeconds - (hours * SECONDS_IN_HOUR) - (minutes * SECONDS_IN_MINUTE);
std::cout << "Elapsed time for " << i << ": " << hours << "h, " << minutes << "m, " << seconds << "s\n";
}
}
int main()
{
constexpr unsigned int MAX_NUM_CHECK = 10000000;
for (int i = 1000000; i <= MAX_NUM_CHECK; i++)
{
auto start = std::chrono::system_clock::now();
int digitIndex = 1;
// Simplifying this to move to the next i in the loop early:
// if i % 2 then the last digit is a 0, 2, 4, 6, or 8 and is therefore divisible by 2
// if i % 5 then the last digit is 0 or 5 and is therefore divisible by 5
if (i % 2 == 0 || i % 5 == 0)
{
std::cout << i << " not prime" << '\n';
auto end = std::chrono::system_clock::now();
std::chrono::duration<double> elapsed_seconds = end - start;
PrintTime(elapsed_seconds.count(), i);
continue;
}
bool isPrime = true;
int divisionNumber = 3;
int floorNum = floorSqrt(i);
while (divisionNumber <= floorNum && isPrime)
{
if (divisionNumber % 5 == 0)
{
divisionNumber += 2;
continue;
}
int number = 0;
int totalDigits = getTotalDigits(i);
// This section does the division necessary to iterate through each digit of the 1 to N number
// Example: Think of dividing 124 into 123456 on paper and how you would iterate through that process
while (digitIndex <= totalDigits)
{
number *= 10;
number += getIndexDigit(i, digitIndex);
number %= divisionNumber;
digitIndex++;
}
if (number == 0)
{
isPrime = false;
break;
}
divisionNumber += 2;
}
if (isPrime)
std::cout << "N = " << i << " is prime." << '\n';
else
std::cout << i << " not prime" << '\n';
auto end = std::chrono::system_clock::now();
std::chrono::duration<double> elapsed_seconds = end - start;
PrintTime(elapsed_seconds.count(), i);
}
}
Its nice to see you are working on the same question I pondered few months ago.
Please refer to question posted in Math Stackexchange for better resources.
TL-DR,
The number you are looking for is called SmarandachePrime.
As per your code, it seems you are dividing with every number that is not a multiple of 2,5. To optimize you can actually check for n = 6k+1 ( 𝑘 ∈ ℕ ).
unfortunately, it is still not a better approach with respect to the number you are dealing with.
The better approach is to use primality test screening to find probable prime numbers in the sequence and then check whether they are prime or not. These tests take a less time ~(O(k log3n)) to check whether a number is prime or not, using mathematical fundamentals, compared to division.
there are several libraries that provide functions for primality check.
for python, you can use gmpy2 library, which uses Miller-Rabin Primality test to find probable primes.
I recommend you to further read about different Primality tests here.
I believe you are missing one very important check, and it's the division by 3:
A number can be divided by 3 is the sum of the numbers can be divided by 3, and your number consists of all numbers from 1 to N.
The sum of all numbers from 1 to N equals:
N * (N+1) / 2
This means that, if N or N+1 can be divided by 3, then your number cannot be prime.
So before you do anything, check MOD(N,3) and MOD(N+1,3). If either one of them equals zero, you can't have a prime number.
We have given a binary String of length n,we can cyclically shift this string any number of times.Let X be the decimal representation of string s. Find the greatest power of 2 with which X can be divisible with, if it can be divisible with arbitrarily large power print "-1".For the result, you are required to print a single integer denoting the maximum power of 2 by which X can be divisible with.
ex:
Input:
0011
Output:
2
Explanation:We can cyclically shift the string 2 times to get "1100" which is divisible by 2^2 hence the answer is 2.
Here is my solution .. however it is giving me tle on most of the test cases and wrong answer on some of the test cases..
int highestpower(int n)
{
return (n & (~(n - 1)));
}
int findnum(string s)
{
int value = 0;
int p=0;
for(int i = s.length()-1;i>=0;i--)
{
value = value+pow(2,p)*(s[i]-'0');
p++;
}
return value;
}
int maximumPower(string s) {
int ans = 0;
for(int i=0;i<s.length();i++)
{
int num = findnum(s.substr(i)+s.substr(0,i));
ans = max(ans,highestpower(num));
}
return ans/2;
}
how can I solve this answer?Thanks..
I have some difficulty to understand the logic of your code. In practice, it failed on about all cases I have tested.
Moreover, it seems quite over-complicated. It is enough to count the number of consecutive zeros. We just have to pay attention that
this calculation must be performed in a cyclic way. For example, if s == 00100, the count number is 4, as after shifting, we get 10000. One simple way to handle this cyclicity is to concatenate the string s2 = s+s = 0010000100 and then to count the maximum number of consecutive zeros in the obtained string s2. In addition, we must pay attention that the input string is not composed of zeros only.
In the following code, I compared your code (maximumPower) with mine (maximumPower_new), on several different inputs.
Result:
0011 : 2 new: 2
0100010 : 4 new: 3
00100 : 8 new: 4
The code:
#include <iostream>
#include <string>
#include <cmath>
#include <algorithm>
int highestpower(int n)
{
return (n & (~(n - 1)));
}
int findnum(const std::string& s)
{
int value = 0;
int p=0;
for(int i = s.length()-1;i>=0;i--)
{
value = value+pow(2,p)*(s[i]-'0');
p++;
}
return value;
}
int maximumPower(const std::string& s) {
int ans = 0;
for(int i = 0; i < s.length(); i++)
{
int num = findnum(s.substr(i)+s.substr(0,i));
ans = std::max(ans,highestpower(num));
}
return ans/2;
}
int maximumPower_new (const std::string& s) {
int n = s.length();
if (n == 0) return -1;
std::string s2 = s + s;
int count = 0;
int count_max = 0;
for (auto c: s2) {
if (c == '0') {
count ++;
} else {
count_max = std::max(count, count_max);
count = 0;
}
}
count_max = std::max(count, count_max);
if (count_max >= n) return -1;
else return count_max;
}
int main() {
for (std::string s: {"0011", "0100010", "00100"}) {
std::cout << s << " : " << maximumPower(s) << " new: " << maximumPower_new(s) << "\n";
}
}
Basically i want to write a function that takes values from 0 to 6 and gives back a random assortment such as 2,3,4,5,0,1,6. Here is the code that i came up with. However the problem is that the integer prev (meaning previous) does not store all the old values of r (random number) and thus some values end up being repeated. How might i fix this?
int s(int b)
{
// b is 7
int h = b-1;
int prev = -1;// to store the previous r value
srand(time(0));
for (int i = 0; i < b; i++)
{
int r = rand()%(h - 0 + 1) + 0;
if (r != prev)
{
cout << r << endl;
prev = r;
}
else if (r == prev)
{
s(b);
}
}
return 0;
}
From the comments, this sounds more like a homework problem than a practical problem because you said "No arrays allowed". But I suppose it is an interesting problem.
Here's some code, in Java with only loops, if statements, and with no arrays, as required.
It outputs a random permutation of the set 0, 1, ..., N, shuffled with the Fisher-Yates algorithm.
void printRandom(int N) {
long used = 0;
for (int i = 0; i < N; i++) {
int randomIndex = ThreadLocalRandom.current().nextInt(N - Long.bitCount(used));
for (int j = 0; j < N; j++) {
if ((used & (1L << j)) == 0) {
if (randomIndex-- == 0) {
System.out.print(j + " ");
used = used | (1L << j);
break;
}
}
}
}
}
It is unfortunately limited to the size of a long on your system :)
I think the best way to solve this problem is by using an aux funtion that stores in a variable all the numbers printed until the moment, check if the new number is in the used numbers variable, if not add it to the variable (you can use strings? I know that they are arrays of char's but maybe you can)
Something like this:
function aux(int b, char *variables_printed, int iterations_left)
if (b = 0) then print variables_printed
else
int n = generate_random_number() %b
while (n in variables_printed)
n= (n+random_number) % b
variables_printed += n
aux(b, variables_printed, iterations_left-1)
And your other function:
function s(b)
if b < 0 return 0
else
char *variables_to_print
aux(b, variables_to_print, b)
If you can not use strings, you can do it with long as konsolas said.
I am obliged to write program that will show minimal quantity of Fibonacci numbers which sum is equal to k number which user write into program.
So for example user writes 1070
And the program will show 987 89 -5 -1 (because all of these numbers are Fibonacci numbers (you can add or substract to get desired sum).
I have problem to think about solution to this problem.
I wrote code to get all Fibonacci numbers from 0 to 46.
#include <iostream>
unsigned long long Fibbo[47];
void preapare()
{
Fibbo[0] = 0;
Fibbo[1] = 1;
i = 2;
while (i<47)
{
Fibbo[i] = Fibbo[i - 1] + Fibbo[i - 2];
i++;
}
}
int main()
{
preapare();
int k=0;
std::cin >> k;
}
I will be glad for any help. You can use Fibonacci Number as many times as you will. For example you can do 1+1+1+1+1
#include <iostream>
#include <vector>
#include <stdlib.h>
using namespace std;
void prepare( vector<int> & fibos ) { ... } // along the lines we see in OPs code
int findClosest( int n, vector<int> & fibos ){ // return Fibonacci number closest to n
int fi = 0;
for( std::vector<int>::iterator it = fibos.begin() ; it != fibos.end(); ++it){
if( abs(*it - n) < abs(fi - n) ){
fi = *it;
}
}
return fi;
}
// The algorithm:
// lookup closest Fi, add "signed" to terms
// reduce n, adjust sign and call recursively
void sum( int n, vector<int> & fibos, vector<int> & terms, int sign = 1 ){
if( n == 0 ) return;
int fi = findClosest( n, fibos );
terms.push_back( sign*fi );
sum( abs(n - fi), fibos, terms, n - fi > 0 ? sign : -sign );
}
int main() {
vector<int> fibos;
prepare( fibos );
vector<int> terms;
int n = 1070;
sum( n, fibos, terms );
cout << n << " =";
for( std::vector<int>::iterator it = terms.begin() ; it != terms.end(); ++it){
cout << " " << *it;
}
cout << endl;
return 0;
}
For user input:
Before the while loop, you need to get the input, save it to a variable, and use that in your while condition. Something like this:
#include <iostream>
...
int k_val = 0;
std::cout << "Enter the value for k: ";
std::cin >> k_val;
...
Your prepare() fixed:
unsigned long long* preapare(const size_t n) {
unsigned long long* Fibbo = new unsigned long long[n];
Fibbo[0] = 0;
Fibbo[1] = 1;
for(size_t i = 2; i < n; ++i) {
Fibbo[i] = Fibbo[i - 1] + Fibbo[i - 2];
}
return Fibbo;
}
then you'll have to memory manage the returned array;
I've done this, the code will solve the grater fibonacci numbers to be add so the result will be the specified number:
fiboMax return a sequence with the fibonacci number less to the input number.
getiboSec return a sequence with the bigger fibonacci numbers that sum the input number.
'''
Created on 15/01/2014
'''
def fiboMax(num):
ret = [0,1]
a = 1
b = 2
while b < num:
ret.append(b)
tmp = a + b
a = b
b = tmp
return ret
def getFiboSec(num):
fiboSec = []
fiboNumbers = fiboMax(num)
i = len(fiboNumbers) - 1
while num > 0 and i > 0:
fiboNum = fiboNumbers[i]
while(fiboNum <= num):
fiboSec.append(fiboNum)
num -= fiboNum
i -= 1
return fiboSec
if __name__ == '__main__':
num = 20
print(fiboMax(num))
print(getFiboSec(num))
output:
figonnaci number to 20 = [0, 1, 2, 3, 5, 8, 13]
fibonacci number that sum 20 = [13, 5, 2]
It doesnt work with negative numbers, anyway its solving the problem but not the efficient way.
You are accessing the indexes -2 and -1 in your loop. Your variable i should start at 2. You also try to add values to pointers. None of your values in Fibbo[] are set.
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.
Given an integer n, display the first
triangle number having at least n
divisors.
Sample Input: 5
Output 28
Input Constraints: 1<=n<=320
I was obviously able to do this question, but I used a naive algorithm:
Get n.
Find triangle numbers and check their number of factors using the mod operator.
But the challenge was to show the output within 4 seconds of input. On high inputs like 190 and above it took almost 15-16 seconds. Then I tried to put the triangle numbers and their number of factors in a 2d array first and then get the input from the user and search the array. But somehow I couldn't do it: I got a lot of processor faults. Please try doing it with this method and paste the code. Or if there are any better ways, please tell me.
Here's a hint:
The number of divisors according to the Divisor function is the product of the power of each prime factor plus 1. For example, let's consider the exponential prime representation of 28:
28 = 22 * 30 * 50 * 71 * 110...
The product of each exponent plus one is: (2+1)*(0+1)*(0+1)*(1+1)*(0+1)... = 6, and sure enough, 28 has 6 divisors.
Now, consider that the nth triangular number can be computed in closed form as n(n+1)/2. We can multiply numbers written in the exponential prime form simply by adding up the exponents at each position. Dividing by two just means decrementing the exponent on the two's place.
Do you see where I'm going with this?
Well, you don't go into a lot of detail about what you did, but I can give you an optimization that can be used, if you didn't think of it...
If you're using the straightforward method of trying to find factors of a number n, by using the mod operator, you don't need to check all the numbers < n. That obviously would take n comparisons...you can just go up to floor(sqrt(n)). For each factor you find, just divide n by that number, and you'll get the conjugate value, and not need to find it manually.
For example: say n is 15.
We loop, and try 1 first. Yep, the mod checks out, so it's a factor. We divide n by the factor to get the conjugate value, so we do (15 / 1) = 15...so 15 is a factor.
We try 2 next. Nope. Then 3. Yep, which also gives us (15 / 3) = 5.
And we're done, because 4 is > floor(sqrt(n)). Quick!
If you didn't think of it, that might be something you could leverage to improve your times...overall you go from O(n) to O(sqrt (n)) which is pretty good (though for numbers this small, constants may still weigh heavily.)
I was in a programming competition way back in school where there was some similar question with a run time limit. the team that "solved" it did as follows:
1) solve it with a brute force slow method.
2) write a program to just print out the answer (you found using the slow method), which will run sub second.
I thought this was bogus, but they won.
see Triangular numbers: a(n) = C(n+1,2) = n(n+1)/2 = 0+1+2+...+n. (Formerly M2535 N1002)
then pick the language you want implement it in, see this:
"... Python
import math
def diminishing_returns(val, scale):
if val < 0:
return -diminishing_returns(-val, scale)
mult = val / float(scale)
trinum = (math.sqrt(8.0 * mult + 1.0) - 1.0) / 2.0
return trinum * scale
..."
First, create table with two columns: Triangle_Number Count_of_Factors.
Second, derive from this a table with the same columns, but consisting only of the 320 rows of the lowest triangle number with a distinct number of factors.
Perform your speedy lookup to the second table.
If you solved the problem, you should be able to access the thread on Project Euler in which people post their (some very efficient) solutions.
If you're going to copy and paste a problem, please cite the source (unless it was your teacher who stole it); and I second Wouter van Niferick's comment.
Well, at least you got a good professor. Performance is important.
Since you have a program that can do the job, you can precalculate all of the answers for 1 .. 320.
Store them in an array, then simply subscript into the array to get the answer. That will be very fast.
Compile with care, winner of worst code of the year :D
#include <iostream>
bool isPrime( unsigned long long number ){
if( number != 2 && number % 2 == 0 )
return false;
for( int i = 3;
i < static_cast<unsigned long long>
( sqrt(static_cast<double>(number)) + 1 )
; i += 2 ){
if( number % i == 0 )
return false;
}
return true;
}
unsigned int p;
unsigned long long primes[1024];
void initPrimes(){
primes[0] = 2;
primes[1] = 3;
unsigned long long number = 5;
for( unsigned int i = 2; i < 1024; i++ ){
while( !isPrime(number) )
number += 2;
primes[i] = number;
number += 2;
}
return;
}
unsigned long long nextPrime(){
unsigned int ret = p;
p++;
return primes[ret];
}
unsigned long long numOfDivs( unsigned long long number ){
p = 0;
std::vector<unsigned long long> v;
unsigned long long prime = nextPrime(), divs = 1, i = 0;
while( number >= prime ){
i = 0;
while( number % prime == 0 ){
number /= prime;
i++;
}
if( i )
v.push_back( i );
prime = nextPrime();
}
for( unsigned n = 0; n < v.size(); n++ )
divs *= (v[n] + 1);
return divs;
}
unsigned long long nextTriNumber(){
static unsigned long long triNumber = 1, next = 2;
unsigned long long retTri = triNumber;
triNumber += next;
next++;
return retTri;
}
int main()
{
initPrimes();
unsigned long long n = nextTriNumber();
unsigned long long divs = 500;
while( numOfDivs(n) <= divs )
n = nextTriNumber();
std::cout << n;
std::cin.get();
}
def first_triangle_number_with_over_N_divisors(N):
n = 4
primes = [2, 3]
fact = [None, None, {2:1}, {3:1}]
def num_divisors (x):
num = 1
for mul in fact[x].values():
num *= (mul+1)
return num
while True:
factn = {}
for p in primes:
if p > n//2: break
r = n // p
if r * p == n:
factn = fact[r].copy()
factn[p] = factn.get(p,0) + 1
if len(factn)==0:
primes.append(n)
factn[n] = 1
fact.append(factn)
(x, y) = (n-1, n//2) if n % 2 == 0 else (n, (n-1)//2)
numdiv = num_divisors(x) * num_divisors(y)
if numdiv >= N:
print('Triangle number %d: %d divisors'
%(x*y, numdiv))
break
n += 1
>>> first_triangle_number_with_over_N_divisors(500)
Triangle number 76576500: 576 divisors
Dude here is ur code, go have a look. It calculates the first number that has divisors greater than 500.
void main() {
long long divisors = 0;
long long nat_num = 0;
long long tri_num = 0;
int tri_sqrt = 0;
while (1) {
divisors = 0;
nat_num++;
tri_num = nat_num + tri_num;
tri_sqrt = floor(sqrt((double)tri_num));
long long i = 0;
for ( i=tri_sqrt; i>=1; i--) {
long long remainder = tri_num % i;
if ( remainder == 0 && tri_num == 1 ) {
divisors++;
}
else if (remainder == 0 && tri_num != 1) {
divisors++;
divisors++;
}
}
if (divisors >100) {
cout <<"No. of divisors: "<<divisors<<endl<<tri_num<<endl;
}
if (divisors > 500)
break;
}
cout<<"Final Result: "<<tri_num<<endl;
system("pause");
}
Boojum's answer motivated me to write this little program. It seems to work well, although it does use a brute force method of computing primes. It's neat how all the natural numbers can be broken down into prime number components.
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <iomanip>
#include <vector>
//////////////////////////////////////////////////////////////////////////////
typedef std::vector<size_t> uint_vector;
//////////////////////////////////////////////////////////////////////////////
// add a prime number to primes[]
void
primeAdd(uint_vector& primes)
{
size_t n;
if (primes.empty())
{
primes.push_back(2);
return;
}
for (n = *(--primes.end()) + 1; ; ++n)
{
// n is even -> not prime
if ((n & 1) == 0) continue;
// look for a divisor in [2,n)
for (size_t i = 2; i < n; ++i)
{
if ((n % i) == 0) continue;
}
// found a prime
break;
}
primes.push_back(n);
}
//////////////////////////////////////////////////////////////////////////////
void
primeFactorize(size_t n, uint_vector& primes, uint_vector& f)
{
f.clear();
for (size_t i = 0; n > 1; ++i)
{
while (primes.size() <= i) primeAdd(primes);
while (f.size() <= i) f.push_back(0);
while ((n % primes[i]) == 0)
{
++f[i];
n /= primes[i];
}
}
}
//////////////////////////////////////////////////////////////////////////////
int
main(int argc, char** argv)
{
// allow specifying number of TN's to be evaluated
size_t lim = 1000;
if (argc > 1)
{
lim = atoi(argv[1]);
}
if (lim == 0) lim = 1000;
// prime numbers
uint_vector primes;
// factors of (n), (n + 1)
uint_vector* f = new uint_vector();
uint_vector* f1 = new uint_vector();
// sum vector
uint_vector sum;
// prime factorize (n)
size_t n = 1;
primeFactorize(n, primes, *f);
// iterate over triangle-numbers
for (; n <= lim; ++n)
{
// prime factorize (n + 1)
primeFactorize(n + 1, primes, *f1);
while (f->size() < f1->size()) f->push_back(0);
while (f1->size() < f->size()) f1->push_back(0);
size_t numTerms = f->size();
// compute prime factors for (n * (n + 1) / 2)
sum.clear();
size_t i;
for (i = 0; i < numTerms; ++i)
{
sum.push_back((*f)[i] + (*f1)[i]);
}
--sum[0];
size_t numFactors = 1, tn = 1;
for (i = 0; i < numTerms; ++i)
{
size_t exp = sum[i];
numFactors *= (exp + 1);
while (exp-- != 0) tn *= primes[i];
}
std::cout
<< n << ". Triangle number "
<< tn << " has " << numFactors << " factors."
<< std::endl;
// prepare for next iteration
f->clear();
uint_vector* tmp = f;
f = f1;
f1 = tmp;
}
delete f;
delete f1;
return 0;
}