HDU 5150 Sum Sum Sum 素数

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;

}