# Prime numbers in a given range using STL | Set 2

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 = 100
Output : 53 59 61 67 71 73 79 83 89 97

Input : start = 900 end = 1000
Output : 907 911 919 929 937 941 947 953 967 971 977 983 991 997
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Idea is to use Sieve of Eratosthenes as a subroutine. We have discussed one implementation in Prime numbers in a given range using STL | Set 1

1. Find primes in the range from 0 to end and store it in a vector
2. Find the index of element less than start value using binary search. We use lower_bound() in STL.
3. Erase elements from the beginning of the vector to that index. We use vector erase()

Viola! The vector contains primes ranging from start to end.

 `// C++ program to print all primes ` `// in a range using Sieve of Eratosthenes ` `#include ` `#include ` `#include ` `#include ` `using` `namespace` `std; ` ` `  `#define all(v) v.begin(), v.end() ` `typedef` `unsigned ``long` `long` `int` `ulli; ` ` `  `vector 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 = ``false``; ` `    ``prime = ``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 ans; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``if` `(prime[i]) ` `            ``ans.push_back(i); ` `    ``return` `ans; ` `} ` ` `  `vector sieveRange(ulli start, ulli end) ` `{ ` `    ``// find primes from [0..end] range ` `    ``vector ans = sieve(end); ` ` `  `    ``// Find index of first prime greater than or ` `    ``// equal to start ` `    ``// O(sqrt(n)loglog(n)) ` `    ``int` `lower_bound_index = lower_bound(all(ans), start) -  ` `                                              ``ans.begin(); ` ` `  `    ``// Remove all elements smaller than start. ` `    ``// O(logn) ` `    ``ans.erase(ans.begin(), ans.begin() + lower_bound_index); ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver Program to test above function ` `int` `main(``void``) ` `{ ` `    ``ulli start = 50; ` `    ``ulli end = 100; ` `    ``vector ans = sieveRange(start, end); ` `    ``for` `(``auto` `i : ans) ` `        ``cout << i << ``' '``; ` `    ``return` `0; ` `} `

Output:
```53 59 61 67 71 73 79 83 89 97
```

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