Sum Sum Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 290    Accepted Submission(s): 194

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   难点是把：primes[1]=1;  

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <string.h>
using namespace std;

const int MAXN = 1001;
bool flag[MAXN];
int primes[MAXN], pi;
void GetPrime_1()
{
    int i, j;
    pi = 0;
    memset(flag, false, sizeof(flag));
    for (i = 2; i < MAXN; i++)
        if (!flag[i])
        {
            primes[i] = 1;//素数标识为1
            for (j = i; j < MAXN; j += i)
                flag[j] = true;
        }
}

int main()
{
    memset(primes,0,sizeof(primes));
    GetPrime_1();
    primes[1]=1;
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        long long ans=0;
        int a;
        for(int i=0;i<n;i++)
        {
            cin>>a;
            if(primes[a]==1)
                ans+=a;
        }
        cout<<ans<<endl;
    }
    return 0;
}