【CUGBACM15级BC第24场 A】hdu 5150 Sum Sum Sum

Sum Sum Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1356    Accepted Submission(s): 817

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

#include<bits/stdc++.h>
#include <ctime>
#define N 1050
using namespace std;
typedef long long ll;

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()
{
    std::ios::sync_with_stdio(false);
    int n, t;
    init();
    while (cin >> n)
    {
        long long ans = 0;
        for (int i = 0; i < n; i++)
        {
            cin >> t;
            if (prime[t])
            {
                ans += t;
            }
        }
        cout << ans << endl;
    }
    return 0;
}