Sum Sum Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

We call a positive number X P-number if there is not a positive number that is less than X and the greatest common divisor of these two numbers is bigger than 1.
Now you are given a sequence of integers. You task is to calculate the sum of P-numbers of the sequence.

Input

There are several test cases. 
In each test case:
The first line contains a integer N(1≤N≤1000). The second line contains N integers. Each integer is between 1 and 1000.

Output

For each test case, output the sum of P-numbers of the sequence.

Sample Input

3
5 6 7
1
10

Sample Output

12
0

Source

BestCoder Round #24

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=2e5+10,M=1e6+10,inf=1e9+10,mod=1e9+7;
const ll INF=1e18+10;
int prime(int n)
{
    if(n<=1)
    return 0;
    if(n==2)
    return 1;
    if(n%2==0)
    return 0;
    int k, upperBound=n/2;
    for(k=3; k<=upperBound; k+=2)
    {
        upperBound=n/k;
        if(n%k==0)
            return 0;
    }
    return 1;
}
int flag[1010];
int main()
{
    int n,x;
    for(int i=2;i<=1000;i++)
    flag[i]=prime(i);
    flag[1]=1;
    while(~scanf("%d",&n))
    {
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            if(flag[x])
            ans+=x;
        }
        printf("%d\n",ans);
    }
    return 0;
}