PRIME1 - Prime Generator


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.

Example

Input:
2
1 10
3 5

Output:
2
3
5
7

3
5
Warning: large Input/Output data, be careful with certain languages (though most should be OK if the algorithm is well designed)

Information

After cluster change, please consider PRINT as a more challenging problem.

hide comments
rinka1: 2019-12-05 21:31:06

i have got time limit exceeded!!

fiodhy_an: 2019-12-04 01:11:56

Please help i got TLE over and over, what shoud i do?

monk_: 2019-12-03 11:43:56

use sieve of eratosthenes algorithm

abhihacker: 2019-12-02 16:05:43

Getting time limit exceeded

sandipan2224: 2019-12-02 08:32:16

Last edit: 2019-12-02 08:32:49
sandipan2224: 2019-12-02 08:31:36

Getting TLE again!!! Does simple sieve work?

bugtoaster: 2019-12-02 04:10:58

why its wrong
<!---
#include<iostream>
#include <sstream>
#include <vector>
using namespace std;

bool isPrime(int n){
if (n <= 1){
return false;
}

for (int i = 2; i < n; i++){
if (n % i == 0){
return false;
}else{
return true;
}
}

}

int main(){
int m=0,n=0,i,tmp,vPosition;
vector<int> v;
string linInt;
getline(cin, linInt);
istringstream modifyStr(linInt);

while (modifyStr >> tmp) {
v.push_back(tmp);
}
for (vector<int>::iterator it = v.begin(); it != v.end(); it++) {

vPosition = std::distance(v.begin(), it);
if(vPosition==0){
m=*it;
}else if(vPosition == 1){
n=*it;
}
}

if(m>0 && n>0){

if(m>n){

int abstr = m - n;
for( i=0; i <= abstr; i++){
int cm = n + i;
isPrime(cm)? cout << cm<<endl : cout;

}
}else if(n>m){
int abstr = n - m;
for(i=0; i <= abstr; i++){
int cm = m + i;
isPrime(cm)? cout << cm<<endl : cout;
}
}else{
return 0;
}
}

}

--->

krishp: 2019-12-01 04:45:33

You arent supposed to use a brute force approach but rather something like segmented sieve or sieve of atkin.

cody_khon: 2019-11-28 17:02:58

i have got time limit exceeded??

swapnilsah710: 2019-11-18 16:47:33

i am getting time limit exceeded plz help!!!!


Added by:Adam Dzedzej
Date:2004-05-01
Time limit:6s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: NODEJS PERL6