I am making a random number generator. It asks how many digits the user wants to be in the number. for example it they enter 2 it will generate random numbers between 10 and 99. I have made the generator but my issue is that the numbers are not unique.
Here is my code. I am not sure why it is not generating unique number. I thought srand(time(null)) would do it.
void TargetGen::randomNumberGen()
{
srand (time(NULL));
if (intLength == 1)
{
for (int i = 0; i< intQuantity; i++)
{
int min = 1;
int max = 9;
int number1 = rand();
if (intQuantity > max)
{
intQuantity = max;
}
cout << number1 % max + min << "\t";
}
}
else if (intLength == 2)
{
for (int i = 0; i<intQuantity; i++)
{
int min = 10;
int max = 90;
int number1 = rand();
if (intQuantity > max)
{
intQuantity = max;
}
cout << number1 % max + min << "\t";
}
}
if (intLength == 3)
{
for (int i = 0; i<intQuantity; i++)
{
int min = 100;
int max = 900;
int number1 = rand();
if (intQuantity > max)
{
intQuantity = max;
}
cout << number1 % max + min << "\t";
}
}
else if (intLength == 4)
{
for (int i = 0; i<intQuantity; i++)
{
int min = 1000;
int max = 9000;
int number1 = rand();
if (intQuantity > max)
{
intQuantity = max;
}
cout << number1 % max + min << "\t";
}
}
if (intLength == 5)
{
for (int i = 0; i<intQuantity; i++)
{
int min = 10000;
int max = 90000;
int number1 = rand();
if (intQuantity > max)
{
intQuantity = max;
}
cout << number1 % max + min << "\t";
}
}
else if (intLength == 6)
{
for (int i = 0; i<intQuantity; i++)
{
int min = 100000;
int max = 900000;
int number1 = rand();
if (intQuantity > max)
{
intQuantity = max;
}
cout << number1 % max + min << "\t";
}
}
if (intLength == 7)
{
for (int i = 0; i<intQuantity; i++)
{
int min = 1000000;
int max = 9000000;
int number1 = rand();
if (intQuantity > max)
{
intQuantity = max;
}
cout << number1 % max + min << "\t";
}
}
else if (intLength == 8)
{
for (int i = 0; i <intQuantity; i++)
{
int min = 10000000;
int max = 89999999;
int number1 = rand();
if (intQuantity > max)
{
intQuantity = max;
}
cout << number1 % max + min << "\t";
}
}
if (intLength == 9)
{
for (int i = 0; i < intQuantity; i++)
{
int min = 100000000;
int max = 900000000;
int number1 = rand();
if (intQuantity > max)
{
intQuantity = max;
}
cout << number1 % max + min << "\t";
}
}
}
Okay so I thought I figured out a way to do this without arrays but It isn't working before I switch to the fisher yates method. Can someone tell me why this isn't working? It is supposed to essentially take the random number put that into variable numGen. Then in variable b = to numgen. Just to hold what numGen used to be so when the loop goes through and generates another random number it will compare it to what the old number is and if it is not equal to it, then it will output it. If it is equal to the old number than rather than outputting it, it will deincrement i so that it will run through the loop without skipping over the number entirely. However, when I do this is infinitely loops. And I am not sure why.
if (intLength == 1)
{
for (int i = 0; i< intQuantity; ++i)
{
int min = 1;
int max = 9;
int number1 = rand();
int numGen = number1 % max + min;
if (intQuantity > max)
{
intQuantity = max;
}
for (int k = 0; k < 1; k++)
{
cout << numGen << "\t";
int b = numGen;
}
int b = numGen;
if (b != numGen )
{
cout << numGen << "\t";
}
else
{
i--;
}
}
}
Everyone has interesting expectations for random numbers -- apparently, you expect random numbers to be unique! If you use any good random number generator, your random numbers will never be guaranteed to be unique.
To make this most obvious, if you wanted to generate random numbers in the range [1, 2], and you were to generate two numbers, you would (normally expect to) get one of the following four possibilities with equal probability:
1, 2
2, 1
1, 1
2, 2
It does not make sense to ask a good random number generator to generate the first two, but not the last two.
Now, take a second to think what to expect if you asked to generate three numbers in the same range... 1, 2, then what??
Uniqueness, therefore, is not, and will not be a property of a random number generator.
Your specific problem may require uniqueness, though. In this case, you need to do some additional work to ensure uniqueness.
One way is to keep a tab on which numbers are already picked. You can keep them in a set, and re-pick if you get one you got earlier. However, this is effective only if you pick a small set of numbers compared to your range; if you pick most of the range, the end of the process gets ineffective.
If the number count you are going to pick corresponds to most of the range, then using an array of the range, and the using a good shuffling algorithm to shuffle the numbers around is a better solution. (The Fisher-Yates shuffle should do the trick.)
Hint 0:
Use Quadratic residue from number theory; an integer q is called a quadratic residue modulo p if it is congruent to a perfect square modulo p; i.e., if there exists an integer x such that:
x2 ≡ q (mod p)
Hint 1:
Theorem: Assuming p is a prime number, the quadratic residue of x is unique as long as 2x < p. For example:
02 ≡ 0 (mod 13)
12 ≡ 1 (mod 13)
22 ≡ 4 (mod 13)
32 ≡ 9 (mod 13)
42 ≡ 3 (mod 13)
52 ≡ 12 (mod 13)
62 ≡ 10 (mod 13)
Hint 2:
Theorem: Assuming p is a prime number such that p ≡ 3 (mod 4), not only x2%p (i.e the quadratic residue) is unique for 2x < p but p - x2%p is also unique for 2x>p. For example:
02%11 = 0
12%11 = 1
22%11 = 4
32%11 = 9
42%11 = 5
52%11 = 3
11 - 62%11 = 8
11 - 72%11 = 6
11 - 82%11 = 2
11 - 92%11 = 7
11 - 102%11 = 10
Thus, this method provides us with a perfect 1-to-1 permutation on the integers less than p, where p can be any prime such that p ≡ 3 (mod 4).
Hint 3:
unsigned int UniqueRandomMapping(unsigned int x)
{
const unsigned int p = 11; //any prime number satisfying p ≡ 3 (mod 4)
unsigned int r = ((unsigned long long) x * x) % p;
if (x <= p / 2) return r;
else return p - r;
}
I didn't worry about the bad input numbers (e.g. out of the range).
Remarks
For 32-bit integers, you may choose the largest prime number such that p ≡ 3 (mod 4) which is less than 232 which is 4294967291.
Even though, this method gives you a 1-to-1 mapping for generating random number, it suffers from the clustering issue.
To improve the randomness of the aforementioned method, combine it with
other unique random mapping methods such as XOR operator.
I'll assume you can come up with a way to figure out how many numbers you want to use. It's pretty simple, since a user input of 2 goes to 10-99, 3 is 100-999, etc.
If you want to come up with your own implementation of unique, randomly generated numbers, check out these links.
http://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle
Here is a very similar implementation: https://stackoverflow.com/a/196065/2142219
In essence, you're creating an array of X integers, all set to the value of their index. You randomly select an index between 0 and MAX, taking the value at this index and swapping it with the max value. MAX is then decremented by 1 and you can repeat it by randomly selecting an index between 0 and MAX - 1.
This gives you a random array of 0-999 integers with no duplicates.
Here are two possible approaches to generating unique random numbers in a range.
Keep track of which numbers you have already generated using std::set, and throw away and regenerate numbers as long as they are already in the set. This approach is not recommended if you want to generate a large number of random numbers, due to the birthday paradox.
Generate all numbers in your given range, take a random permutation of them, and output however many the user wants.
Standard random generators would never generate unique numbers, in this case they would Not be independent.
To generate unique numbers you have to:
Save all number generated and compare new one with old ones, if there is coincidence - regenerate.
or
Use random_shuffle function: http://en.cppreference.com/w/cpp/algorithm/random_shuffle to get all sequence in advance.
Firstly, srand()/rand() commonly have a period of 2^32, which means that after calling srand(), rand() will internally iterate over distinct integers during the first 2^32 calls to rand(). Still, rand() may well return a result with less than 32 bits: such as an int between 0 and RAND_MAX where RAND_MAX is 2^31-1 or 2^15-1, so you may see repeated results as the caller of rand(). You probably read about the period though, or somebody's comment made with awareness of that, and somehow it's been mistaken as uniqueness....
Secondly, given any call to rand() generates a number far larger than you want, and you're doing this...
number1 % max
The result of "number1 % max" is in the range 0 <= N <= max, but the random number itself may have been any multiple of max greater than that. In other words, two distinct random numbers that differ by a multiple of max still produce the same result for number1 % max in your program.
To get distinct random numbers within a range, you could prepopulate a std::vector with all the numbers, then std::shuffle them.
Related
The question is to find the number of interesting numbers lying between two numbers. By the interesting number, they mean that the product of its digits is divisible by the sum of its digits.
For example: 459 => product = 4 * 5 * 9 = 180, and sum = 4 + 5 + 9 = 18; 180 % 18 == 0, hence it is an interesting number.
My solution for this problem is having run time error and time complexity of O(n2).
#include<iostream>
using namespace std;
int main(){
int x,y,p=1,s=0,count=0,r;
cout<<"enter two numbers"<<endl;
cin>>x>>y;
for(int i=x;i<=y;i++)
{
r=0;
while(i>1)
{
r=i%10;
s+=r;
p*=r;
i/=10;
}
if(p%s==0)
{
count++;
}
}
cout<<"count of interesting numbers are"<<count<<endl;
return 0;
}
If s is zero then if(p%s==0) will produce a divide by zero error.
Inside your for loop you modify the value of i to 0 or 1, this will mean the for loop never completes and will continuously check 1 and 2.
You also don't reinitialise p and s for each iteration of the for loop so will produce the wrong answer anyway. In general limit the scope of variables to where they are actually needed as this helps to avoid this type of bug.
Something like this should fix these problems:
#include <iostream>
int main()
{
std::cout << "enter two numbers\n";
int begin;
int end;
std::cin >> begin >> end;
int count = 0;
for (int number = begin; number <= end; number++) {
int sum = 0;
int product = 1;
int value = number;
while (value != 0) {
int digit = value % 10;
sum += digit;
product *= digit;
value /= 10;
}
if (sum != 0 && product % sum == 0) {
count++;
}
}
std::cout << "count of interesting numbers are " << count << "\n";
return 0;
}
I'd guess the contest is trying to get you to do something more efficient than this, for example after calculating the sum and product for 1234 to find the sum for 1235 you just need to add one and for the product you can divide by 4 then multiply by 5.
Given a natural number n (1 <= n <= 500000), please output the summation of all its proper divisors.
Definition: A proper divisor of a natural number is the divisor that is strictly less than the number.
e.g. number 20 has 5 proper divisors: 1, 2, 4, 5, 10, and the divisor summation is: 1 + 2 + 4 + 5 + 10 = 22.
<<--This is a challenge i am trying to do and i am using recursion
int find_sum(std::vector <int> nums,long int sum,int num_now,long int j)
{
if(j<nums[num_now])
{
if(nums[num_now]%j==0)
{
sum=sum+j;
}
return find_sum(nums,sum,num_now,j+1);
}
else
{
return sum;
}
}
sum is the sum of all divisors,nums is the vector i stored number in,num_now is current member in vector,int j is 1 i use it to search for dividers,sadly using this i cant use numbers like 500000 it give's me error,is there any better way to do it or have i made a mistake somewhere.
--Thank you for your time
Here is a recursive way to solve your problem:
int find_sum(int x, int i)
{
if(i == 0)
return 0;
if(x % i == 0)
return i + find_sum(x, (i-1));
return find_sum(x, (i-1));
}
You need to call find_sum(N, N-1); in order to find sum of dividers of N (i must be less than given N because of strict inequality).
In your case it would be find_sum(20, 19);
e.g. my function returns:
71086 for N = 50000
22 for N = 20
0 for N = 1
I don't see the reason why you need to use recursion for solving this problem. I would prefer a more staightforward way to solve it.
long CalculateSumOfDivisors(int number)
{
long sum = 0;
for(int i=1; i<number; i++)
{
// If the remainder of num/i is zero
// then i divides num. So we add it to the
// current sum.
if(number%i==0)
{
sum+=i;
}
}
return sum;
}
Furthermore, we could write a more optimal algorithm, if we note the following:
Let that we have a number n and d is the smallest divisor of n that is greater of 1. (Apparently if the number n is a prime number there is any such a divisor). Then the larget divisor of n is the number n/d.
Based on this we can formulate a more optimal algorithm.
long CalculateSumOfDivisors(int number)
{
int smallestDivisor = FindSmallestDivisor(number);
if(smallestDivisor==1) return 1;
long sum = smallestDivisor;
// Calculate the possible greatest divisor.
int possibleGreatestDivisor = (int)floor(number/smallestDivisor);
for(int i=smallestDivisor+1; i<=possibleGreatestDivisor; i++)
{
if(number%i==0)
{
sum+=i;
}
}
return sum;
}
int FindSmallestDivisor(int number)
{
int smallestDivisor = 1;
for(int i=2; i<number; i++)
{
if(number%i==0)
{
smallestDivisor = i;
break;
}
}
return smallestDivisor;
}
I tried writing code with main function asking user to give the it wants to get sum of. here is the code , hope it helps.
#include<iostream>
using namespace std;
int Sum(int min, int max, int &val, int &sum){
if(min >= max)
return 0;
for ( ; min < max; min++){
if ( val%min == 0){
sum += min + val/min;
return Sum(++min,val/min, val,sum);
}
}
return 0;
}
int main(){
int s=1;
int val;
cout <<"Enter Val to sum:";
cin >> val;
Sum(2,val,val,s);
cout <<"Sum is :"<<s<<endl;
return 0;
}
Here Sum function is used recursively and passed parameters as shown in the code.
Hope it helps.
I don't think you should use recursion.
Instead start by looping from 1..N-1
When you find a divisor adjust the end value for the loop. Example if 2 is a divisor then you know N/2 is also a divisor. And just as important you know there can be no further divisors in the range ]N/2:N[. Likewise if 3 is a divisor then you know N/3 is also a divisor and you know there are no more divisors in the range ]N/3:N[.
Following that concept you can reduced the number of loops significantly for most numbers.
Something like:
long find_sum(int num)
{
long sum = 0;
int max = num;
int i = 1;
while(i < max)
{
if(num % i == 0)
{
sum += i; // Add i to the sum
max = num / i; // Decrement max for performance
if (max != i && max != num)
{
sum += max; // Add max when max isn't equal i
}
}
i++;
}
return sum;
}
Example:
num = 10
sum = 0
i = 1 -> sum = 1, max = 10
i = 2 -> sum = 1+2+5, max = 5
i = 3 -> sum = 1+2+5, max = 5
i = 4 -> sum = 1+2+5, max = 5
i = 5 -> return 8 (1+2+5)
num = 64
sum = 0
i = 1 -> sum = 1, max = 64
i = 2 -> sum = 1+2+32, max = 32
i = 3 -> sum = 1+2+32, max = 32
i = 4 -> sum = 1+2+32+4+16, max = 16
i = 5 -> sum = 1+2+32+4+16, max = 16
i = 6 -> sum = 1+2+32+4+16, max = 16
i = 7 -> sum = 1+2+32+4+16, max = 16
i = 8 -> sum = 1+2+32+4+16+8, max = 8
i = 9 -> return (1+2+32+4+16+8)
The number of loops are kept down by changing max whenever a new divisor is found.
I'm trying to solve this exercise http://main.edu.pl/en/archive/amppz/2014/dzi and I have no idea how to improve perfomance of my code. Problems occure when program have to handle over 500,000 unique numbers(up to 2,000,000 as in description). Then it took 1-8s to loop over all this numbers. Tests I have used are from http://main.edu.pl/en/user.phtml?op=tests&c=52014&task=1263, and I testing it by command
program.exe < data.in > result.out
Description:
You are given a sequence of n integer a1, a2, ... an. You should determine the number of such ordered pairs(i, j), that i, j equeals(1, ..., n), i != j and ai is divisor of aj.
The first line of input contains one integer n(1 <= n <= 2000000)
The second line contains a sequence of n integers a1, a2, ..., an(1 <= ai <= 2000000).
In the first and only line of output should contain one integer, denoting the number of pairs sought.
For the input data:
5
2 4 5 2 6
the correct answer is: 6
Explanation: There are 6 pars: (1, 2) = 4/2, (1, 4) = 2/2, (1, 5) = 6/2, (4, 1) = 2/2, (4, 2) = 4/2, (4, 5) = 6/2.
For example:
- with 2M in total numbers and 635k unique numbers, there is 345mln iterations in total
- with 2M in total numbers and 2mln unqiue numbers, there is 1885mln iterations in total
#include <iostream>
#include <math.h>
#include <algorithm>
#include <time.h>
#define COUNT_SAME(count) (count - 1) * count
int main(int argc, char **argv) {
std::ios_base::sync_with_stdio(0);
int n; // Total numbers
scanf("%d", &n);
clock_t start, finish;
double duration;
int minVal = 2000000;
long long *countVect = new long long[2000001]; // 1-2,000,000; Here I'm counting duplicates
unsigned long long counter = 0;
unsigned long long operations = 0;
int tmp;
int duplicates = 0;
for (int i = 0; i < n; i++) {
scanf("%d", &tmp);
if (countVect[tmp] > 0) { // Not best way, but works
++countVect[tmp];
++duplicates;
} else {
if (minVal > tmp)
minVal = tmp;
countVect[tmp] = 1;
}
}
start = clock();
int valueJ;
int sqrtValue, valueIJ;
int j;
for (int i = 2000000; i > 0; --i) {
if (countVect[i] > 0) { // Not all fields are setted up
if (countVect[i] > 1)
counter += COUNT_SAME(countVect[i]); // Sum same values
sqrtValue = sqrt(i);
for (j = minVal; j <= sqrtValue; ++j) {
if (i % j == 0) {
valueIJ = i / j;
if (valueIJ != i && countVect[valueIJ] > 0 && valueIJ > sqrtValue)
counter += countVect[i] * countVect[valueIJ];
if (i != j && countVect[j] > 0)
counter += countVect[i] * countVect[j];
}
++operations;
}
}
}
finish = clock();
duration = (double)(finish - start) / CLOCKS_PER_SEC;
printf("Loops time: %2.3f", duration);
std::cout << "s\n";
std::cout << "\n\nCounter: " << counter << "\n";
std::cout << "Total operations: " << operations;
std::cout << "\nDuplicates: " << duplicates << "/" << n;
return 0;
}
I know, I shouldn't sort the array at beginning, but I have no idea how to make it in better way.
Any tips will be great, thanks!
Here is improved algorithm - 2M unique numbers within 0.5s. Thanks to #PJTraill!
#include <iostream>
#include <math.h>
#include <algorithm>
#include <time.h>
#define COUNT_SAME(count) (count - 1) * count
int main(int argc, char **argv) {
std::ios_base::sync_with_stdio(0);
int n; // Total numbers
scanf("%d", &n);
clock_t start, finish;
double duration;
int maxVal = 0;
long long *countVect = new long long[2000001]; // 1-2,000,000; Here I'm counting duplicates
unsigned long long counter = 0;
unsigned long long operations = 0;
int tmp;
int duplicates = 0;
for (int i = 0; i < n; i++) {
scanf("%d", &tmp);
if (countVect[tmp] > 0) { // Not best way, but works
++countVect[tmp];
++duplicates;
} else {
if (maxVal < tmp)
maxVal = tmp;
countVect[tmp] = 1;
}
}
start = clock();
int j;
int jCounter = 1;
for (int i = 0; i <= maxVal; ++i) {
if (countVect[i] > 0) { // Not all fields are setted up
if (countVect[i] > 1)
counter += COUNT_SAME(countVect[i]); // Sum same values
j = i * ++jCounter;
while (j <= maxVal) {
if (countVect[j] > 0)
counter += countVect[i] * countVect[j];
j = i * ++jCounter;
++operations;
}
jCounter = 1;
}
}
finish = clock();
duration = (double)(finish - start) / CLOCKS_PER_SEC;
printf("Loops time: %2.3f", duration);
std::cout << "s\n";
std::cout << "\n\nCounter: " << counter << "\n";
std::cout << "Total operations: " << operations;
std::cout << "\nDuplicates: " << duplicates << "/" << n;
return 0;
}
I expect the following to work a lot faster than the OP’s algorithm (optimisations oblique):
(The type of values and frequencies should be 32-bit unsigned, counts 64-bit – promote before calculating a count, if your language would not.)
Read the number of values, N.
Read each value v, adding one to its frequency freq[v] (no need to store it).
(freq[MAX] (or MAX+1) can be statically allocated for probably optimal initialisation to all 0)
Calculate the number of pairs involving 1 from freq[1] and the number of values.
For every i in 2..MAX (with freq[i] > 0):
Calculate the number of pairs (i,i) from freq[i].
For every multiple m of i in 2m..MAX:
(Use m as the loop counter and increment it, rather than multiplying)
Calculate the number of pairs (i,m) from freq[i] and freq[m].
(if freq[i] = 1, one can omit the (i,i) calculation and perform a variant of the loop optimised for freq[i] = 1)
(One can perform the previous (outer) loop from 2..MAX/2, and then from MAX/2+1..MAX omitting the processing of multiples)
The number of pairs (i,i) = freq[i]C2 = ( freq[i] * (freq[i] - 1) ) / 2 .
The number of pairs (i,j) = freq[i] * freq[j] for i ≠ j.
This avoids sorting, sqrt and division.
Other optimisations
One can store the distinct values, and scan that array instead (the order does not matter); the gain or loss due to this depends on the density of the values in 1..MAX.
If the maximum frequency is < 216, which sounds very probable, all products will fit in 32 bits. One could take advantage of this by writing functions with the numeric type as a template, tracking the maximum frequency and then choosing the appropriate instance of the template for the rest. This costs N*(compare+branch) and may gain by performing D2 multiplications with 32 bits instead of 64, where D is the number of distinct values. I see no easy way to deduce that 32 bits suffice for the total, apart from N < 216.
If parallelising this for n processors, one could let different processors process different residues modulo n.
I considered keeping track of the number of even values, to avoid a scan of half the frequencies, but I think that for most datasets within the given parameters that would yield little advantage.
Ok, I am not going to write your whole algorithm for you, but it can definitely be done faster. So i guess this is what you need to get going:
So you have your list sorted, so there are a lot of assumptions you can make from this. Take for instance the highest value. It wont have any multiples. The highest value that does, will highest value divided by two.
There is also one other very usefull fact here. A multiple of a multiple is also a multiple. (Still following? ;)). Take for instance the list [2 4 12]. Now you've found (4,12) as a multiple pair. If you now also find (2,4), then you can deduce that 12 is also a multiple of 2.
And since you only have to count the pairs, you can just keep a count for each number how many multiples it has, and add that when you see that number as a multiple itself.
This means that it is probably best to iterate your sorted list backwards, and look for divisors instead.
And maybe store it in some way that goes like
[ (three 2's ), (two 5's), ...]
ie. store how often a number occurs. Once again, you don't have to keep track of it's id, since you only need to give them the total number of pairs.
Storing your list this way helps you, because all the 2's are going to have the same amount of multiples. So calculate once and then multiply.
Here's the problem.
Write the given number N, as sum of the given numbers, using only additioning and subtracting.
Here's an example:
N = 20
Integers = 8, 15, 2, 9, 10
20 = 8 + 15 - 2 + 9 - 10.
Here's my idea;
First idea was to use brute force, alternating plus and minus. First I calculate the number of combinations and its 2^k (where k is the nubmer of integers), because I can alternate only minus and plus. Then I run through all numbers from 1 to 2^k and I convert it to binary form. And for any 1 I use plus and for any 0 I use minus. You'll get it easier with an example (using the above example).
The number of combinations is: 2^k = 2^5 = 32.
Now I run through all numbers from 1 to 32.
So i get: 1=00001, that means: -8-15-2-9+10 = -24 This is false so I go on.
2 = 00010, which means: -8-15-2+9-10 = -26. Also false.
This method works good, but when the number of integers is too big it takes too long.
Here's my code in C++:
#include <iostream>
#include <cmath>
using namespace std;
int convertToBinary(int number) {
int remainder;
int binNumber = 0;
int i = 1;
while(number!=0)
{
remainder=number%2;
binNumber=binNumber + (i*remainder);
number=number/2;
i=i*10;
}
return binNumber;
}
int main()
{
int N, numberOfIntegers, Combinations, Binary, Remainder, Sum;
cin >> N >> numberOfIntegers;
int Integers[numberOfIntegers];
for(int i = 0; i<numberOfIntegers; i++)
{
cin >>Integers[i];
}
Combinations = pow(2.00, numberOfIntegers);
for(int i = Combinations-1; i>=Combinations/2; i--) // I use half of the combinations, because 10100 will compute the same sum as 01011, but in with opposite sign.
{
Sum = 0;
Binary = convertToBinary(i);
for(int j = 0; Binary!=0; j++)
{
Remainder = Binary%10;
Binary = Binary/10;
if(Remainder==1)
{
Sum += Integers[numberOfIntegers-1-j];
}
else
{
Sum -= Integers[numberOfIntegers-1-j];
}
}
if(N == abs(Sum))
{
Binary = convertToBinary(i);
for(int j = 0; Binary!=0; j++)
{
Remainder = Binary%10;
Binary = Binary/10;
if(Sum>0)
{
if(Remainder==1)
{
cout << "+" << Integers[numberOfIntegers-1-j];
}
else
{
cout << "-" << Integers[numberOfIntegers-1-j];
}
}
else
{
if(Remainder==1)
{
cout << "-" << Integers[numberOfIntegers-1-j];
}
else
{
cout << "+" << Integers[numberOfIntegers-1-j];
}
}
}
break;
}
}
return 0;
}
Since this is typical homework, I'm not going to give the complete answer. But consider this:
K = +a[1] - a[2] - a[3] + a[4]
can be rewritten as
a[0] = K
a[0] + a[2] + a[3] = a[1] + a[4]
You now have normal subset sums on both sides.
So what you are worried about is you complexity .
Lets analyse what optimisations can be done.
Given n numbers in a[n] and target Value T;
And it is sure one combination of adding and subtracting gives you T ;
So Sigma(m*a[k]) =T where( m =(-1 or 1) and 0 >= k >= n-1 )
This just means ..
It can written as
(sum of Some numbers in array) = (Sum of remaining numbers in array) + T
Like in your case..
8+15-2+9-10=20 can be written as
8+15+9= 20+10+2
So Sum of all numbers including target = 64 // we can cal that .. :)
So half of it is 32 as
Which if further written as 20+(somthing)=32
which is 12 (2+10) in this case.
Your problem can be reduced to Finding the numbers in an array whose sum is 12 in this case
So your problem now can be reduced as find the combination of numbers whose sum is k (which you can calculate as described above k=12 .) For Which the complexity is O(log (n )) n as size of array , Keep in mind that you have to sort array and use binary search based algo for getting O(log(n)).
So as complexity can be made from O(2^n) to O((N+1)logN)as sorting included.
This takes static input as you have provided and i have written using core java
public static void main(String[] args) {
System.out.println("Enter number");
Scanner sc = new Scanner(System.in);
int total = 0;
while (sc.hasNext()) {
int[] array = new int[5] ;
for(int m=0;m<array.length;m++){
array[m] = sc.nextInt();
}
int res =array[0];
for(int i=0;i<array.length-1;i++){
if((array[i]%2)==1){
res = res - array[i+1];
}
else{
res =res+array[i+1];
}
}
System.out.println(res);
}
}
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;
}