Skip to main content
Rollback to Revision 3
Source Link
Mast
  • 13.8k
  • 12
  • 57
  • 127
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
#include <cmath>
#include <vector>
#include<iostream>
using namespace std;

void simplesieve(int limit, vector<int>&prime);


void segmentedsieve() {
    int T, M, K;

    cin >> T;
        **vector<int> prime;
        simplesieve(32000, prime);**

    for (int t = 0; t < T; t++) {
        cin >> M >> K;
        const int limit = floor(sqrt(K)) + 1;
        vector<int> prime;
        simplesieve(K, prime);

        bool *mark = new bool[K - M];
        memset(mark, true, sizeof(bool)*(K-M));
        
        vector<int>::iterator it;
     
        for (it = prime.begin(); *it<limit; it++) {

            int lolim = floor(M / (*it))*(*it);
            if (lolim < M)
                lolim = lolim + (*it);

            for (int j = lolim; j < K; j += (*it))
            {   if(j!=*it)
                mark[j - M] = false;
            }


        }

        for (int i = M; i < K; i++) {
            if (mark[i - M] == true)
                cout << i << "\n";
        }

        delete[] mark;
    }

    
}



void simplesieve( int limit, vector<int>&prime) {
    
    bool *mark=new bool[limit + 1];
    
    memset(mark, true, sizeof(bool)*(limit+1));
    for (int p = 2; p*p < limit; p++) {
        if (mark[p] == true) {
            for (int i = p * 2; i < limit; i += p)
                mark[i] = false;
        }
    }
    for (int p = 2; p < limit; p++) {
        if (mark[p] == true)
        {
            prime.push_back(p);
        
        }

    }

}


int main()
{
    int n = 100;
    segmentedsieve();
    return 0;
}
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
#include <cmath>
#include <vector>
#include<iostream>
using namespace std;

void simplesieve(int limit, vector<int>&prime);


void segmentedsieve() {
    int T, M, K;

    cin >> T;
        **vector<int> prime;
        simplesieve(32000, prime);**

    for (int t = 0; t < T; t++) {
        cin >> M >> K;
        const int limit = floor(sqrt(K)) + 1;
        

        bool *mark = new bool[K - M];
        memset(mark, true, sizeof(bool)*(K-M));
        
        vector<int>::iterator it;
     
        for (it = prime.begin(); *it<limit; it++) {

            int lolim = floor(M / (*it))*(*it);
            if (lolim < M)
                lolim = lolim + (*it);

            for (int j = lolim; j < K; j += (*it))
            {   if(j!=*it)
                mark[j - M] = false;
            }


        }

        for (int i = M; i < K; i++) {
            if (mark[i - M] == true)
                cout << i << "\n";
        }

        delete[] mark;
    }

    
}



void simplesieve( int limit, vector<int>&prime) {
    
    bool *mark=new bool[limit + 1];
    
    memset(mark, true, sizeof(bool)*(limit+1));
    for (int p = 2; p*p < limit; p++) {
        if (mark[p] == true) {
            for (int i = p * 2; i < limit; i += p)
                mark[i] = false;
        }
    }
    for (int p = 2; p < limit; p++) {
        if (mark[p] == true)
        {
            prime.push_back(p);
        
        }

    }

}


int main()
{
    int n = 100;
    segmentedsieve();
    return 0;
}
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
#include <cmath>
#include <vector>
#include<iostream>
using namespace std;

void simplesieve(int limit, vector<int>&prime);


void segmentedsieve() {
    int T, M, K;

    cin >> T;


    for (int t = 0; t < T; t++) {
        cin >> M >> K;
        const int limit = floor(sqrt(K)) + 1;
        vector<int> prime;
        simplesieve(K, prime);

        bool *mark = new bool[K - M];
        memset(mark, true, sizeof(bool)*(K-M));
        
        vector<int>::iterator it;
     
        for (it = prime.begin(); *it<limit; it++) {

            int lolim = floor(M / (*it))*(*it);
            if (lolim < M)
                lolim = lolim + (*it);

            for (int j = lolim; j < K; j += (*it))
            {   if(j!=*it)
                mark[j - M] = false;
            }


        }

        for (int i = M; i < K; i++) {
            if (mark[i - M] == true)
                cout << i << "\n";
        }

        delete[] mark;
    }

    
}



void simplesieve( int limit, vector<int>&prime) {
    
    bool *mark=new bool[limit + 1];
    
    memset(mark, true, sizeof(bool)*(limit+1));
    for (int p = 2; p*p < limit; p++) {
        if (mark[p] == true) {
            for (int i = p * 2; i < limit; i += p)
                mark[i] = false;
        }
    }
    for (int p = 2; p < limit; p++) {
        if (mark[p] == true)
        {
            prime.push_back(p);
        
        }

    }

}


int main()
{
    int n = 100;
    segmentedsieve();
    return 0;
}
added 16 characters in body
Source Link
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
#include <cmath>
#include <vector>
#include<iostream>
using namespace std;

void simplesieve(int limit, vector<int>&prime);


void segmentedsieve() {
    int T, M, K;

    cin >> T;
        **vector<int> prime;
        simplesieve(32000, prime);**

    for (int t = 0; t < T; t++) {
        cin >> M >> K;
        const int limit = floor(sqrt(K)) + 1;
        vector<int> prime;
        simplesieve(K, prime);

        bool *mark = new bool[K - M];
        memset(mark, true, sizeof(bool)*(K-M));
        
        vector<int>::iterator it;
     
        for (it = prime.begin(); *it<limit; it++) {

            int lolim = floor(M / (*it))*(*it);
            if (lolim < M)
                lolim = lolim + (*it);

            for (int j = lolim; j < K; j += (*it))
            {   if(j!=*it)
                mark[j - M] = false;
            }


        }

        for (int i = M; i < K; i++) {
            if (mark[i - M] == true)
                cout << i << "\n";
        }

        delete[] mark;
    }

    
}



void simplesieve( int limit, vector<int>&prime) {
    
    bool *mark=new bool[limit + 1];
    
    memset(mark, true, sizeof(bool)*(limit+1));
    for (int p = 2; p*p < limit; p++) {
        if (mark[p] == true) {
            for (int i = p * 2; i < limit; i += p)
                mark[i] = false;
        }
    }
    for (int p = 2; p < limit; p++) {
        if (mark[p] == true)
        {
            prime.push_back(p);
        
        }

    }

}


int main()
{
    int n = 100;
    segmentedsieve();
    return 0;
}
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
#include <cmath>
#include <vector>
#include<iostream>
using namespace std;

void simplesieve(int limit, vector<int>&prime);


void segmentedsieve() {
    int T, M, K;

    cin >> T;


    for (int t = 0; t < T; t++) {
        cin >> M >> K;
        const int limit = floor(sqrt(K)) + 1;
        vector<int> prime;
        simplesieve(K, prime);

        bool *mark = new bool[K - M];
        memset(mark, true, sizeof(bool)*(K-M));
        
        vector<int>::iterator it;
     
        for (it = prime.begin(); *it<limit; it++) {

            int lolim = floor(M / (*it))*(*it);
            if (lolim < M)
                lolim = lolim + (*it);

            for (int j = lolim; j < K; j += (*it))
            {   if(j!=*it)
                mark[j - M] = false;
            }


        }

        for (int i = M; i < K; i++) {
            if (mark[i - M] == true)
                cout << i << "\n";
        }

        delete[] mark;
    }

    
}



void simplesieve( int limit, vector<int>&prime) {
    
    bool *mark=new bool[limit + 1];
    
    memset(mark, true, sizeof(bool)*(limit+1));
    for (int p = 2; p*p < limit; p++) {
        if (mark[p] == true) {
            for (int i = p * 2; i < limit; i += p)
                mark[i] = false;
        }
    }
    for (int p = 2; p < limit; p++) {
        if (mark[p] == true)
        {
            prime.push_back(p);
        
        }

    }

}


int main()
{
    int n = 100;
    segmentedsieve();
    return 0;
}
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
#include <cmath>
#include <vector>
#include<iostream>
using namespace std;

void simplesieve(int limit, vector<int>&prime);


void segmentedsieve() {
    int T, M, K;

    cin >> T;
        **vector<int> prime;
        simplesieve(32000, prime);**

    for (int t = 0; t < T; t++) {
        cin >> M >> K;
        const int limit = floor(sqrt(K)) + 1;
        

        bool *mark = new bool[K - M];
        memset(mark, true, sizeof(bool)*(K-M));
        
        vector<int>::iterator it;
     
        for (it = prime.begin(); *it<limit; it++) {

            int lolim = floor(M / (*it))*(*it);
            if (lolim < M)
                lolim = lolim + (*it);

            for (int j = lolim; j < K; j += (*it))
            {   if(j!=*it)
                mark[j - M] = false;
            }


        }

        for (int i = M; i < K; i++) {
            if (mark[i - M] == true)
                cout << i << "\n";
        }

        delete[] mark;
    }

    
}



void simplesieve( int limit, vector<int>&prime) {
    
    bool *mark=new bool[limit + 1];
    
    memset(mark, true, sizeof(bool)*(limit+1));
    for (int p = 2; p*p < limit; p++) {
        if (mark[p] == true) {
            for (int i = p * 2; i < limit; i += p)
                mark[i] = false;
        }
    }
    for (int p = 2; p < limit; p++) {
        if (mark[p] == true)
        {
            prime.push_back(p);
        
        }

    }

}


int main()
{
    int n = 100;
    segmentedsieve();
    return 0;
}
Tweeted twitter.com/StackCodeReview/status/1045191645891899392
deleted 6 characters in body; edited tags; edited title
Source Link
200_success
  • 145.6k
  • 22
  • 191
  • 481

prime Prime number segmented sieve

iI tried to solve the SPOJ prime number problem below. theThe code produced correct resultresults, but Time limit exceeded and also memory need around 970M which is not efficient(?)

what makes my code slow and howHow can iI optimize this code under a time constraint of around few seconds? thanks!

problem:
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.

Problem

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.

prime number segmented sieve

i tried to solve SPOJ prime number problem. the code produced correct result, but Time limit exceeded and also memory need around 970M which is not efficient(?)

what makes my code slow and how can i optimize this code under time constraint around few seconds? thanks!

problem:
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.

Prime number segmented sieve

I tried to solve the SPOJ prime number problem below. The code produced correct results, but Time limit exceeded and also memory need around 970M which is not efficient(?)

How can I optimize this code under a time constraint of around few seconds?

Problem

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.

deleted 67 characters in body
Source Link
Loading
Source Link
Loading