P5723 Pocket of Prime

Description

Little A has a pocket of primes, where can includes any prime. He always starts at 2 and enumerates if it's a prime. If it's, he will put this number into the pocket. But the load capability of the pocket is the sum of the numbers. The pocket's capability is limited; it cannot load the prime after the sum of primes over L(1<=L<=100000). After given L, asks how many primes can the pocket take? You need to output these primes from small to large and output the capability; there are lines between numbers.

Input format

None

Output format

None

Example

Input
100
Output

2
3
5
7
11
13
17
19
23
9

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Code

#include<iostream>
#include<cstring>
using namespace std;

int prime(){
    int n;
    int c = 0;
    cin >> n;
    int tmp = n;
    bool a[n+1];
    memset(a,true,sizeof(a));
    for(int i=2;i*i<=n;i++){
        if(a[i]==true){
            for(int p=i*i;p<=n;p+=i){
                a[p]=false;
            }
        }
    }
    for(int i=2;i<=n;i++){
        if(a[i]==true){
            tmp -= i;
            if(tmp<0)
                break;
            c++;
            cout << i << endl;
        }
    }
    cout << c << endl;
    return 0;
}

What I learned?

The algorithm to sift prime.
Sieve of Eratosthenes

// C++ program to print all primes smaller than or equal to 
// n using Sieve of Eratosthenes 
#include <bits/stdc++.h> 
using namespace std; 
  
void SieveOfEratosthenes(int n) 
{ 
    // Create a boolean array "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. 
    bool prime[n+1]; 
    memset(prime, true, sizeof(prime)); 
  
    for (int p=2; p*p<=n; p++) 
    { 
        // If prime[p] is not changed, then it is a prime 
        if (prime[p] == true) 
        { 
            // Update all multiples of p greater than or  
            // equal to the square of it 
            // numbers which are multiple of p and are 
            // less than p^2 are already been marked.  
            for (int i=p*p; i<=n; i += p) 
                prime[i] = false; 
        } 
    } 
  
    // Print all prime numbers 
    for (int p=2; p<=n; p++) 
       if (prime[p]) 
          cout << p << " "; 
} 
  
// Driver Program to test above function 
int main() 
{ 
    int n = 30; 
    cout << "Following are the prime numbers smaller "
         << " than or equal to " << n << endl; 
    SieveOfEratosthenes(n); 
    return 0; 
} 

Why pp<=n?
The minimum prime factor of a number n must be less than the root n.
Mathematical expression: ab=n, if a<root n, b must > root n.