hdu 5150 Sum Sum Sum（判断素数）

Sum Sum Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1032    Accepted Submission(s): 611

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

题意：给n个数，求其中素数之和

思路：打表素数筛选，判断素数，注意此题中1也要当成素数。 

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define N 1050
int prime[N],notprime[N];
void init()
{
    prime[1]=1;
    for(int i=2;i<N;i++)
    {
        if(notprime[i]) continue;
        prime[i]=1;
        for(int j=i*i;j<N;j+=i)
            notprime[j]=1;
    }
}
int main()
{
    int n,t;
    init();
    while(~scanf("%d",&n))
    {
        long long ans=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&t);
            if(prime[t]) ans+=t;
        }
        printf("%lld\n",ans);
    }
    return 0;
}