Generate all prime numbers between two given numbers. The task is to print prime numbers in that range. The Sieve of Eratosthenes is one of the most efficient ways to find all primes smaller than n where n is smaller than 10 million or so.

Examples:

Input :start = 50 end = 100Output :53 59 61 67 71 73 79 83 89 97Input :start = 900 end = 1000Output :907 911 919 929 937 941 947 953 967 971 977 983 991 997

Idea is to use Sieve of Eratosthenes as a subroutine. Firstly, find primes in the range from 0 to start and store it in a vector. Similarly, find primes in the range from 0 to end and store in another vector. Now take the set difference of two vectors to obtain the required answer. Remove extra zeros if any in the vector.

`// C++ STL program to print all primes ` `// in a range using Sieve of Eratosthenes ` `#include<bits/stdc++.h>` `using` `namespace` `std;` ` ` `typedef` `unsigned ` `long` `long` `int` `ulli;` ` ` `vector<ulli> sieve(ulli n)` `{` ` ` `// Create a boolean vector "prime[0..n]" and` ` ` `// initialize all entries it as true. A value` ` ` `// in prime[i] will finally be false if i is` ` ` `// Not a prime, else true.` ` ` `vector<` `bool` `> prime(n+1,` `true` `);` ` ` ` ` `prime[0] = ` `false` `;` ` ` `prime[1] = ` `false` `;` ` ` `int` `m = ` `sqrt` `(n);` ` ` ` ` `for` `(ulli p=2; p<=m; p++)` ` ` `{` ` ` ` ` `// If prime[p] is not changed, then it` ` ` `// is a prime ` ` ` `if` `(prime[p])` ` ` `{` ` ` `// Update all multiples of p` ` ` `for` `(ulli i=p*2; i<=n; i += p)` ` ` `prime[i] = ` `false` `;` ` ` `}` ` ` `}` ` ` ` ` `// push all the primes into the vector ans` ` ` `vector<ulli> ans;` ` ` `for` `(` `int` `i=0;i<n;i++)` ` ` `if` `(prime[i])` ` ` `ans.push_back(i);` ` ` `return` `ans;` `}` ` ` `// Used to remove zeros from a vector using ` `// library function remove_if()` `bool` `isZero(ulli i)` `{` ` ` `return` `i == 0;` `}` ` ` `vector<ulli> sieveRange(ulli start,ulli end)` `{` ` ` `// find primes from [0..start] range` ` ` `vector<ulli> s1 = sieve(start); ` ` ` ` ` `// find primes from [0..end] range ` ` ` `vector<ulli> s2 = sieve(end); ` ` ` ` ` `vector<ulli> ans(end-start);` ` ` ` ` `// find set difference of two vectors and` ` ` `// push result in vector ans` ` ` `// O(2*(m+n)-1) ` ` ` `set_difference(s2.begin(), s2.end(), s1.begin(), ` ` ` `s2.end(), ans.begin());` ` ` ` ` `// remove extra zeros if any. O(n)` ` ` `vector<ulli>::iterator itr =` ` ` `remove_if(ans.begin(),ans.end(),isZero);` ` ` ` ` `// resize it. // O(n)` ` ` `ans.resize(itr-ans.begin());` ` ` ` ` `return` `ans;` `}` ` ` `// Driver Program to test above function` `int` `main(` `void` `)` `{ ` ` ` `ulli start = 50;` ` ` `ulli end = 100;` ` ` `vector<ulli> ans = sieveRange(start,end);` ` ` `for` `(` `auto` `i:ans)` ` ` `cout<<i<<` `' '` `;` ` ` `return` `0;` `}` |

Output:

53 59 61 67 71 73 79 83 89 97

This article is contributed by **Varun Thakur**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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