HDU5150 Sum Sum Sum【数论筛选法】

题目链接：Sum Sum Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1590 Accepted Submission(s): 931

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

5 6 7

Sample Output

题意简述：

测试数据有多组，每组首先输入正整数n（1≤n≤1000），然后输入n个整数（1≤整数≤1000）。计算其中素数的和。

程序说明：

使用筛选法打表，再计算总和。

C语言程序如下：

/* HDU5150 Sum Sum Sum */
 
#include <stdio.h>
#include <memory.h>
#include <math.h>
 
#define MAXN 1000
 
int notprime[MAXN+1];
 
void esieve(int p[], int n)
{
    int i, j;
 
    memset(p, 0, MAXN+1);
 
    // 筛选
    int max = (int)sqrt(n);
    for(i=2; i<=max; i++) {
        if(p[i] == 0) {
            for(j=i+i; j<=n; j+=i)
                p[j] = 1;
        }
    }
 
}
int main(void)
{
    esieve(notprime, MAXN);
 
    int n, sum, val, i;
 
    while(scanf("%d", &n) != EOF) {
        sum = 0;
 
        for(i=1; i<=n; i++) {
            scanf("%d", &val);
            if(!notprime[val])
                sum += val;
        }
 
        printf("%d\n", sum);
    }
 
    return 0;
}