chore(CPlusPlus): add segmented sieve for range (#535)
Co-authored-by: Ming Tsai <37890026+ming-tsai@users.noreply.github.com>pull/582/head
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/*
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Problem: finding all prime numbers in a range [L,R], where R <= 1e12 and (R-L+1) <= 1e7.
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*/
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#include <bits/stdc++.h>
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using namespace std;
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//function to find prime numbers between L and R
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void segmented_sieve_range(long int L, long int R){
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// generate all primes up to sqrt(R)
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long int e = sqrt(R);
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// vector to store whether ith element is visited or not
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vector<bool> check(e + 1, false);
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//stores all primes till sqrt(R)
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vector<long int> primes;
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for (long int i = 2; i <= e; ++i) {
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if (!check[i]) {
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//i is a prime number
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primes.push_back(i);
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for (long int j = i*i; j <= e; j += i){
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check[j] = true;
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}
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}
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}
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vector<bool> isPrime(R - L + 1, true);
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//edge case
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if(L == 1){
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isPrime[0] = false;
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}
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for (long int i = 0; i < primes.size() ; i++){
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//current Prime
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long int cur = primes[i];
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long int product = cur*cur;
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for(long int start = max(product, (L + cur -1)/product); start <= R ; start += cur){
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isPrime[start - L] = false;
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}
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}
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//Printing all primes in the range L to R
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for(long int i = 0; i < isPrime.size(); i++){
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if(isPrime[i]){
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//current index element is a prime number
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cout<< L + i << " ";
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}
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}
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cout << endl;
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}
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//main starts
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int main()
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{
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long int L, R;
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cin >> L >> R;
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segmented_sieve_range(L, R);
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return 0;
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}
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/*
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Time Complexity - O((R−L+1)log(R) + √R)
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Input:
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30 50
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Output:
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31 37 41 43 47
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Input:
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10 20
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Output:
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11 13 17 19
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*/
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@ -140,6 +140,7 @@
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- [Missing number](Maths/missing-number.cpp)
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- [Factorial of a number](Maths/factorial.cpp)
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- [Prime-number](Maths/prime-number.cpp)
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- [Segmented Sieve](Maths/segmented-sieve-range.cpp)
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# Recursion
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