This is a program to count the number of divisors for a number, but it is giving one less divisor than there actually is for that number.
#include <stdio.h>
int i = 20;
int divisor;
int total;
int main()
{
for (divisor = 1; divisor <= i; divisor++)
{
if ((i % divisor == 0) && (i != divisor))
{
total = total++;
}
}
printf("%d %d\n", i, total);
return 0;
}
The number 20 has 6 divisors, but the program says that there are 5 divisors.
&& (i != divisor)
means that 20 won't be considered a divisor. If you want it to be considered, ditch that bit of code, and you'll get the whole set, {1, 2, 4, 5, 10, 20}.
Even if you didn't want the number counted as a divisor, you could still ditch that code and just use < instead of <= in the for statement.
And:
total = total++;
is totally unnecessary. It may even be undefined, I'm just too lazy to check at the moment and it's not important since nobody writes code like that for long :-)
Use either:
total = total + 1;
or (better):
total++;
Divisor counting is perhaps simpler and certainly faster than any of these. The key fact to note is that if p is a divisor of n, then so is n/p. Whenever p is not the square root of n, then you get TWO divisors per division test, not one.
int divcount(int n)
{
int i, j, count=0;
for (i=1, j=n; i<j; j = n/++i)
{
if (i*j == n)
count += 2;
}
if (i == j && i*j == n)
++count;
return count;
}
That gets the job done with sqrt(n) divisions, and sqrt(n) multiplications. I choose that because, while j=n/i and another j%i can be done with a single division instruction on most CPUs, I haven't seen compilers pick up on that optimization. Since multiplication is single-clock on modern desktop processors, the i*j == n test is much cheaper than a second division.
PS: If you need a list of divisors, they come up in the loop as i and j values, and perhaps as the i==j==sqrt(n) value at the end, if n is a square.
You have added an extra check && (i != divisor) as explained in given answer.
Here, I wrote the same program using the prime factorisation. This is quick way to find the number of divisor for large number (reference).
// this function return the number of divisor for n.
// if n = (m^a) (n^b) ... where m, n.. are prime factors of n
// then number of divisor d(n) = (a+1)*(b+1)..
int divisorcount(int n){
int divider = 2;
int limit = n/2;
int divisorCount = 1;
int power = 0;
// loop through i=2...n/2
while(divider<=limit){
if(n%divider==0){
// dividing numper using prime factor
// (as smallest number devide a number
// is it's prime factor) and increase the
// power term for prime factor.
power++;
n/=divider;
}
else{
if(power != 0){
// use the prime factor count to calculate
// divisor count.
divisorCount*=(power+1);
}
power = 0;
divider++;
// if n become 1 then we have completed the
// prime factorization of n.
if(n==1){
break;
}
}
}
return divisorCount;
}
Related
Given a range [ L , R ] (both inclusive), I have to tell find the maximum difference between two prime numbers in the given range. There are three answers possible for the given range.
If there is only one distinct prime number in the given range, then maximum difference in this case would be 0.
If there are no prime numbers in the given range, then output for this case would be -1.
Example:
Range: [ 1, 10 ]
The maximum difference between the prime numbers in the given range is 5.
Difference = 7 - 2 = 5
Range: [ 5, 5 ]
There is only one distinct prime number so the maximum difference would be 0.
Range: [ 8 , 10 ]
There is no prime number in the given range so the output for the given range would be -1.
Input Format
The first line of input consists of the number of test cases, T
Next T lines each consists of two space-separated integers, L and R
Constraints
1<= T <=10
2<= L<= R<=10^6
This is my code:
#include <stdio.h>
int isprime(int n)
{
int i,c=0;
for(i=1;i<n;i++)
{
if(n%i==0)
c++;
}
if(c==1)
return 1;
else
return 0;
}
int main()
{
int t; //testnumber
scanf("%d",&t);
for(int k=0;k<t;k++)
{
int l,r; //l=low or floor, r = highest range or ceiling;[l,r]
scanf("%d%d",&l,&r);
int n = r-l; //difference in range
int a[n];
int j=0;
for(int i=l;i<=r;i++)
{
if(isprime(i)==1)
{
a[j] = i;
j++;
}
}
int d = a[j-1]-a[0];
if(j==0)
printf("%d\n",-1);
else
printf("%d\n",d);
}
return 0;
}
When posting on a forum/stack or asking for review, try to name your variables appropriately. Otherwise, it becomes uneasy to follow the code or what is the purpose of which variable.
I wrote the code below hoping you will understand my implementation.
#include <iostream>
#include <math.h>
using namespace std;
void isPrime(int num, int* primeNumber)
{
if (num == 2)
{
*primeNumber = num; //2 is a prime number
return;
}
if (num%2 == 0)
{
return; //num is an even number, so, not prime
}
int limit = sqrt(num);
if (limit*limit == num)
{
return; //num is a square number, so, not prime
}
for (int i = 3; i <= limit; i=i+2)//to find if a number is prime or not, we only have to divide it from 2 to sqrt(num).
{ //i=i+2 skips even number, cause already checked if num is even or not.
if (num % i == 0)
{
return; //`num` is divisible by i, so, not prime
}
}
*primeNumber = num; //no divisible number found. so, num is prime.
}
int main()
{
int testNumber;
cout<< "Enter testNumber: ";
cin>> testNumber;
for (int i = 0; i < testNumber; ++i)
{
int newLow, low, high, lowestPrime = 0, highestPrime = -1;
cin>> low>> high;
newLow = low;
if (low == high)
{
cout<<"0\n";
continue;
}
for (int j = low; j <= high; ++j)//find the lowest prime
{
isPrime(j, &lowestPrime);
if (lowestPrime != 0)//if lowest prime found, lowestprime will no longer be 0
{
//cout<<"lowest prime: "<<lowestPrime<<endl;
newLow = j; //there is no prime number between low...newLow
break;
}
}
for (int j = high; j >= newLow; j--)//find highest prime
{
isPrime(j, &highestPrime);
if (highestPrime != -1)//if highest prime found, highestprime will no longer be -1
{
//cout<<"highest prime: "<<highestPrime<<endl;
break;
}
}
cout<<highestPrime - lowestPrime<<"\n";
}
return 0;
}
This task doesn't require any special algorithm(except checking if number is prime in O(sqrt(N))) to be solved efficiently. Think about prime numbers, what is the frequency of them on some range (for example on range from 1 to 100) what is some "pattern" that appears. Now, if i understood the task correctly you need to find maximal difference of primes on range which is last_prime_on_range - first_prime_on_range, from this and previous observation you can easily devise an efficient algorithm.
Spoiler:
You don't need to check whole range, it would be enough to check from L to L+100
and from R to R-100, obviously if L+100>R you can just go from L to R.
If you want to be sure you can go from L to L+1000 and from R to R-1000 since it doesn't impact time complexity too much.
Also, adding a break; when you find a prime would also solve the problem.
Note that this gap between primes is not guaranteed to be bellow 100/1000 but for given range checking up to 1000 would be enough.
Now if you need to check all primes in range, you should learn about Sieve Of Eratosthenes.
Is there any simple way to make this small program faster? I've made it for an assignment, and it's correct but too slow. The aim of the program is to print the nth pair of primes where the difference between the two is two, given n.
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
bool isPrime(int number) {
for (int i = 3; i <= number/2; i += 2) {
if (!(number%i)) {
return 0;
}
}
return 1;
}
int findNumber(int n) {
int prevPrime, currentNumber = 3;
for (int i = 0; i < n; i++) {
do {
prevPrime = currentNumber;
do {
currentNumber+=2;
} while (!isPrime(currentNumber));
} while (!(currentNumber - 2 == prevPrime));
}
return currentNumber;
}
int main(int argc, char *argv[]) {
int numberin, numberout;
scanf ("%d", &numberin);
numberout = findNumber(numberin);
printf("%d %d\n", numberout - 2, numberout);
return 0;
}
I considered using some kind of array or list that would contain all primes found up until the current number and divide each number by this list instead of all numbers, but we haven't really covered these different data structures yet so I feel I should be able to solve this problem without. I'm just starting with C, but I have some experience in Python and Java.
To find pairs of primes which differ by 2, you only need to find one prime and then add 2 and test if it is also prime.
if (isPrime(x) && isPrime(x+2)) { /* found pair */ }
To find primes the best algorithm is the Sieve of Eratosthenes. You need to build a lookup table up to (N) where N is the maximum number that you can get. You can use the Sieve to get in O(1) if a number is prime. While building the Sieve you can build a list of sorted primes.
If your N is big you can also profit from the fact that a number P is prime iif it doesn't have any prime factors <= SQRT(P) (because if it has a factor > SQRT(N) then it should also have one < SQRT(N)). You can build a Sieve of Eratosthenes with size SQRT(N) to get a list of primes and then test if any of those prime divides P. If none divides P, P is prime.
With this approach you can test numbers up to 1 billion or so relatively fast and with little memory.
Here is an improvement to speed up the loop in isPrime:
bool isPrime(int number) {
for (int i = 3; i * i <= number; i += 2) { // Changed the loop condition
if (!(number%i)) {
return 0;
}
}
return 1;
}
You are calling isPrime more often than necessary. You wrote
currentNummber = 3;
/* ... */
do {
currentNumber+=2;
} while (!isPrime(currentNumber));
...which means that isPrime is called for every odd number. However, when you identified that e.g. 5 is prime, you can already tell that 10, 15, 20 etc. are not going to be prime, so you don't need to test them.
This approach of 'crossing-out' multiples of primes is done when using a sieve filter, see e.g. Sieve of Eratosthenes algorithm in C for an implementation of a sieve filter for primes in C.
Avoid testing ever 3rd candidate
Pairs of primes a, a+2 may only be found a = 6*n + 5. (except pair 3,5).
Why?
a + 0 = 6*n + 5 Maybe a prime
a + 2 = 6*n + 7 Maybe a prime
a + 4 = 6*n + 9 Not a prime when more than 3 as 6*n + 9 is a multiple of 3
So rather than test ever other integer with + 2, test with
a = 5;
loop {
if (isPrime(a) && isPrime(a+2)) PairCount++;
a += 6;
}
Improve loop exit test
Many processors/compilers, when calculating the remainder, will also have available, for nearly "free" CPU time cost, the quotient. YMMV. Use the quotient rather than i <= number/2 or i*i <= number to limit the test loop.
Use of sqrt() has a number of problems: range of double vs. int, exactness, conversion to/from integer. Recommend avoid sqrt() for this task.
Use unsigned for additional range.
bool isPrime(unsigned x) {
// With OP's selective use, the following line is not needed.
// Yet needed for a general purpose `isPrime()`
if (x%2 == 0) return x == 2;
if (x <= 3) return x == 3;
unsigned p = 1;
unsigned quotient, remainder;
do {
p += 2;
remainder = x%p;
if (remainder == 0) return false;
quotient = x/p; // quotient for "free"
} while (p < quotient); // Low cost compare
return true;
}
My Program to calculate the largest prime factor of 600851475143, is stuck and never stops during compilation and execution. Does anyone know why it does not finish execution?
#include <stdio.h> //Edited the #includes(typo) to #include
int main (void)
{
long long int num = 600851475143 ;
long long int factorCount;
long long int bigFactor;
for ( long long int i=1 ; i <= num; i+=2 )// Iterating through all numbers from 2, smaller than or equal to num
{
if ( num % i == 0) // If the number "i" is a factor of num i.e. If "i" perfectly divides num
{
factorCount = 0;
//checking whether a factor "i" , is a prime factor of num
for ( long long int j=2; j <= i ; j++ ) // Iterating through all numbers from 2, smaller than or equal to "i"
{
if ( i % j == 0) // If the number "j" prefectly divides or is a factor of "i"
{
factorCount++; //Add 1 to factorCount
};
};
if ( factorCount == 1 ) // If factorCount has not exceeded 1 i.e., the number "i" is a prime number
{
bigFactor = i;
};
};
};
printf("The largets prime factor of %lli is %lli\n",num,bigFactor );
return 0;
}
I am not sure whether I understood your question .. so you just want to get the biggest prime factor for a certain number? If this is the case, then just do the following:
#include <stdio.h>
#define NUMBER 600851475143
int main (void)
{
long long int num = NUMBER;
long long int factor;
for (factor=2 ; num != 1; factor++) {
if (num % factor == 0) {
num = num / factor;
factor = factor-1;
}
}
printf("The largets prime factor of %lli is %lli\n",NUMBER, factor);
return 0;
}
Why this works: the first prime factor you find is the smallest prime factor of the number; the last prime factor is the biggest. Therefore once you have found a prime factor p, there does not exist a prime factor smaller than p, because otherwise you would have found that smaller prime factor before. Hence your next prime factor is greater equals p.
It finishes its execution, it just takes a lot of time.
You are performing a loop 600851475143 / 2 times or about 300 billion times. If the main loop takes 1ms to executes (but it should take more since there is another inner loop) then it means that time required would be about 9.5 years.
Just be patient and your loop will finish.
Try this:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main (void)
{
long long int num = 600851475143 ;
long long int factorCount;
long long int bigFactor;
for ( long long int i=1 ; i <= sqrt(num); i+=2 )// Iterating through all numbers from 2, smaller than or equal to num
{
if ( num % i == 0) // If the number "i" is a factor of num i.e. If "i" perfectly divides num
{
factorCount = 0;
//checking whether a factor "i" , is a prime factor of num
for ( long long int j=2; j <= i ; j++ ) // Iterating through all numbers from 2, smaller than or equal to "i"
{
if ( i % j == 0) // If the number "j" prefectly divides or is a factor of "i"
{
factorCount++; //Add 1 to factorCount
};
};
if ( factorCount == 1 ) // If factorCount has not exceeded 1 i.e., the number "i" is a prime number
{
bigFactor = i;
};
};
};
printf("The largets prime factor of %lli is %lli\n",num,bigFactor );
return 0;
}
Only calcule the root of 600851475143.
Good Afternoon 24,
The runtime of this code will take very long, very very long. But it should work, only real error is the 's' in
#includes <stdio.h>
so just be patient, you're dealing with very large numbers, iterating through odd numbers has made it much less lengthy than it could be, don
Note: If you are using an online IDE such as cpp.sh or other source, the website will most likely time out. Please use a locally installed IDE.
Well, my best guess would be that it runs perfectly fine but takes too much time.
So what you have to do is optimize your algorithm.
Here are some hints for improving your algorithm:
You just need to iterate until the square root of a number to know whether or not it is a prime.
As you have done in our outer loop (but not in your inner loop), you just have to iterate through odd numbers.
Since you are looking for the highest prime factor, try beginning from the end and once you have reached a prime factor, stop looking.
I think that this should run in a reasonable amount of time.
EDIT: Actually, after reflexion the third point is not that obvious. Of course, it is better than going through all factors but the computation is way slower for large factors ...
This is reasonably efficient:
#include <stdio.h>
#define NUMBER 600851475143
static unsigned long long int gpf(unsigned long long n)
{
unsigned long long d, q;
while (n > 2 && !(n & 1))
n >>= 1;
d = 3;
while (d * d <= n) {
q = n / d;
if (q * d == n)
n = q;
else
d += 2;
}
return n;
}
int main(void)
{
printf("The largest prime factor of %llu is %llu\n",
NUMBER, gpf(NUMBER));
return 0;
}
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main() {
int anz;
scanf("%d", &anz);
time_t start = time(0);
int *primZ = malloc(anz * sizeof(int));
primZ[0] = 2;
int Num = 0;
for (int i = 1, num = 3; i < anz; num += 2) {
for (int j = 1; j < i; j++) {
if (num % primZ[j] == 0) {
num += 2;
j = 0;
}
//this part
if (primZ[j] > i / 2)
break;
}
primZ[i] = num;
i++;
printf("%d ,",num);
}
time_t delta = time(0) - start;
printf("%d", delta);
getchar();
getchar();
return 0;
}
The code works perfectly fine, the question is why. The part if(primZ[j] > i/2) makes the program 2 - 3 times faster. It was actually meant to be if(primZ[j] > num/3) which makes perfect sense because num can only be an odd number. But it is the number of found prime numbers. It makes no sense to me. Please explain.
You check if the prime is composite by checking if it divisible by already found prime numbers. But in doing so you only have to check up to and including the square root of the number because any number larger than that that divides the number will leave a smaller number than the square root of the number.
For example 33 is composite, but you only have to check numbers up to 5 to realize that, you don't need to check it being divisible by 11 because it leaves 3 (33/11=3) which we already checked.
This means that you could improve your algorithm by
for (int j = 1; j < i; j++) {
if( primZ[j]*primZ[j] > num )
break;
if (num % primZ[j] == 0) {
num += 2;
j = 0;
}
}
The reason you can get away with comparing with cutting of at i/2 is due to the distribution of the prime numbers. The prime counting function is approximately i = num/log(num) and then you get that i/2 > sqrt(num).
The reason is that the actual bound is much tighter than num/3 - you could use:
if (primZ[j] > sqrt(num))
The reason for that being that if a prime higher than the square root of num divides num, there must also be a lower prime that does (since the result of such a division must be lower than the square root).
This means that as long as i/2 is higher than sqrt(num), the code will work. What happens is that the number of primes lower than a number grows faster than the square root of that number, meaning that (completely accidentally) i/2 is a safe bound to use.
You can check out how your i value behaves here - they call it pi(x), the number of primes less than x.
It makes sense, since if n has two factors one of them is surely less than or equal to n/2, sense the program found no factors of i in primZ that are less than or equal to i/2 it means there's no factors of i -except 1 of course-.
Sense primZ is sorted in ascending order and j only increases, when primeZ[j] > i/2 it indicates that there's no factors of i in primZ that are less than i/2.
P.S.The point of starting the search is stated in the first part of the for statement num=3 , and the recurring statement num += 2 ensures you only test odd numbers
I've been trying to solve the SPOJ problem of Prime number Generator Algorithm.
Here is the question
Peter wants to generate some prime numbers for his cryptosystem. Help
him! Your task is to generate all prime numbers between two given
numbers!
Input
The input begins with the number t of test cases in a single line
(t<=10). In each of the next t lines there are two numbers m and n (1
<= m <= n <= 1000000000, n-m<=100000) separated by a space.
Output
For every test case print all prime numbers p such that m <= p <= n,
one number per line, test cases separated by an empty line.
It is very easy, but the online judge is showing error, I didn't get what the problem meant by 'test cases' and why that 1000000 range is necessary to use.
Here is my code.
#include<stdio.h>
main()
{
int i, num1, num2, j;
int div = 0;
scanf("%d %d", &num1, &num2);
for(i=num1; i<=num2; i++)
{
for(j=1; j<=i; j++)
{
if(i%j == 0)
{
div++;
}
}
if(div == 2)
{
printf("%d\n", i);
}
div = 0;
}
return 0;
}
I can't comment on the alogirthm and whether the 100000 number range allows optimisations but the reason that your code is invalid is because it doesn't seem to be parsing the input properly. The input will be something like:
2
123123123 123173123
987654321 987653321
That is the first line will give the number of sets of input you will get with each line then being a set of inputs. Your program, at a glance, looks like it is just reading the first line looking for two numbers.
I assume the online judge is just looking for the correct output (and possibly reasonable running time?) so if you correct for the right input it should work no matter what inefficiencies are in your algorithm (as others have started commenting on).
The input begins with the number t of test cases in a single line (t<=10)
you haven't got test cases in your programm.
Its wrong
And sorry for my English
2 - //the number of test cases
1 10 - // numbers n,m
3 5 - // numbers
Your programm will work only in first line.
#include <stdio.h>
#include <math.h>
int main()
{
int test;
scanf("%d",&test);
while(test--)
{
unsigned int low,high,i=0,j=2,k,x=0,y=0,z;
unsigned long int a[200000],b[200000];
scanf("%d",&low);
scanf("%d",&high);
for(i=low;i<=high;i++)
a[x++]=i;
for(i=2;i<=32000;i++)
b[y++]=i;
i=0;
while(b[i]*b[i]<=high)
{
if(b[i]!=0)
{
k=i;
for(;k<y;k+=j)
{
if(k!=i)
{
b[k]=0;
}
}
}
i+=1;j+=1;
}
for(i=0;i<y;i++)
{
if(b[i]!=0 && (b[i]>=low && b[i]<=sqrt(high)))
printf("%d\n",b[i]);
}
int c=0;
for(i=0;i<y;i++)
{
if(b[i]!=0 && (b[i]>=1 && b[i]<=sqrt(high)))
b[c++]=b[i];
}
int m=a[0];
for(i=0;i<c;i++)
{
z=(m/b[i])*b[i];k=z-m;
if(k!=0)
k += b[i];
for(;k<x;)
{
if(a[k]!=0)
{
a[k]=0;
}
k+=b[i];
}
}
for(i=0;i<x;i++)
{
if(a[i]!=0 && (a[i]>=2 && a[i]<=(high)))
printf("%d\n",a[i]);
}
printf("\n");
}
return 0;
}
To find primes between m,n where 1 <= m <= n <= 1000000000, n-m<=100000, you need first to prepare the core primes from 2 to sqrt(1000000000) < 32000. Simple contiguous sieve of Eratosthenes is more than adequate for this. (Having sieved the core bool sieve[] array (a related C code is here), do make a separate array int core_primes[] containing the core primes, condensed from the sieve array, in an easy to use form, since you have more than one offset segment to sieve by them.)
Then, for each given separate segment, just sieve it using the prepared core primes. 100,000 is short enough, and without evens it's only 50,000 odds. You can use one pre-allocated array and adjust the addressing scheme for each new pair m,n. The i-th entry in the array will represent the number o + 2i where o is an odd start of a given segment.
See also:
Is a Recursive-Iterative Method Better than a Purely Iterative Method to find out if a number is prime?
Find n primes after a given prime number, without using any function that checks for primality
offset sieve of Eratoshenes
A word about terminology: this is not a "segmented sieve". That refers to the sieving of successive segments, one after another, updating the core primes list as we go. Here the top limit is known in advance and its square root is very small.
The same core primes are used to sieve each separate offset segment, so this may be better described as an "offset" sieve of Eratosthenes. For each segment being sieved, only the core primes not greater than its top limit's square root need be used of course; but the core primes are not updated while each such offset segment is sieved (updating the core primes is the signature feature of the "segmented" sieve).
For such small numbers you can simply search for all primes between 1 and 1000000000.
Take 62.5 mByte of RAM to create a binary array (one bit for each odd number, because we already know that no even number (except of 2) is a prime).
Set all bits to 0 to indicate that they are primes, than use a Sieve of Eratosthenes to set bits to 1 of all number that are not primes.
Do the sieve once, store the resulting list of numbers.
int num;
bool singleArray[100000];
static unsigned long allArray[1000000];
unsigned long nums[10][2];
unsigned long s;
long n1, n2;
int count = 0;
long intermediate;
scanf("%d", &num);
for(int i = 0; i < num; ++i)
{
scanf("%lu", &n1);
scanf("%lu", &n2);
nums[i][0] = n1;
nums[i][1] = n2;
}
for(int i = 0; i < 100000; ++i)
{
singleArray[i] = true;
}
for(int i = 0; i < num; ++i)
{
s = sqrt(nums[i][1]);
for(unsigned long k = 2; k <= s; ++k)
{
for (unsigned long j = nums[i][0]; j <= nums[i][1]; ++j)
{
intermediate = j - nums[i][0];
if(!singleArray[intermediate])
{
continue;
}
if((j % k == 0 && k != j) || (j == 1))
{
singleArray[intermediate] = false;
}
}
}
for(unsigned long m = nums[i][0]; m <= nums[i][1]; ++m)
{
intermediate = m - nums[i][0];
if(singleArray[intermediate])
{
allArray[count++] = m;
}
}
for(int p = 0; p < (nums[i][1] - nums[i][0]); ++p)
{
singleArray[p] = true;
}
}
for(int n = 0; n < count; ++n)
{
printf("%lu\n", allArray[n]);
}
}
Your upper bound is 10^9. The Sieve of Eratosthenes is O(N loglogN) which is too much for that bound.
Here are a few ideas:
Faster primality tests
The problem with a naive solution where you loop over the range [i, j] and check whether each number is prime is that it takes O(sqrt(N)) to test whether a number is prime which is too much if you deal with several cases.
However, you could try a smarter primality testing algorithm. Miller-Rabin is polynomial in the number of bits of N, and for N <= 10^9, you only need to check a = 2, 7 and 61.
Note that I haven't actually tried this, so I can't guarantee it would work.
Segmented sieve
As #KaustavRay mentioned, you could use a segmented sieve. The underlying idea is that if a number N is composite, then it has a prime divisor that is at most sqrt(N).
We use the Sieve of Eratosthenes algorithm to find the prime numbers below 32,000 (roughly sqrt(10^9)), and then for each number in the range [i, j] check whether there is any prime below 32,000 that divides it.
By the prime number theorem about one in log(N) numbers are prime which is small enough to squeeze in the time limit.
#include <iostream>
using namespace std;
int main() {
// your code here
unsigned long int m,n,i,j;int N;
cin>>N;
for(;N>0;N--)
{
cin>>m>>n;
if(m<3)
switch (n)
{
case 1: cout<<endl;continue;
case 2: cout<<2<<endl;
continue;
default:cout<<2<<endl;m=3;
}
if(m%2==0) m++;
for(i=m;i<=n;i+=2)
{
for(j=3;j<=i/j;j+=2)
if(i%j==0)
{j=0;break;}
if(j)
cout<<i<<endl;
}
cout<<endl;
}return 0;}