欧拉筛

本文介绍了一种改进的质数筛选算法,通过枚举『某个数字』i而非直接枚举质因数,来确保每个合数仅由其最小质因数筛选一次。该算法采用标志数组来标记已知的合数,并利用循环来避免重复筛选。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

对于任意一个合数,我们可以拆成最小质数*某个数字的形式。我们枚举『某个数字』i(区分与埃氏筛法的枚举质因数),然后再从第一个质数开始枚举,进行筛选。
为了保证每个合数只被最小质因数筛选,当我们枚举的质数可以整除『某个数字』时,如果再往大里枚举,枚举的质数就不可能是最小质数了。(挺好正的)

板子

int cntprime = 0;
for (int i = 2; i <= n; i++)
{
    if (!flag[i]) prime[++cntprime] = i;
    for (int j = 1; j <= cntprime && prime[j] * i <= n; j++)
    {
        flag[i * prime[j]] = true;
        if (i % prime[j] == 0)
            break;
    }
}
You and your team have worked tirelessly until you have a sequence a1,a2,…,a2n+1 of positive integers satisfying these properties. 1≤ai≤1018 for all 1≤i≤2n+1 . a1,a2,…,a2n+1 are pairwise distinct. a1=a2&minus;a3+a4&minus;a5+…+a2n&minus;a2n+1 . However, the people you worked with sabotaged you because they wanted to publish this sequence first. They deleted one number from this sequence and shuffled the rest, leaving you with a sequence b1,b2,…,b2n . You have forgotten the sequence a and want to find a way to recover it. If there are many possible sequences, you can output any of them. It can be proven under the constraints of the problem that at least one sequence a exists.    Input Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤104 ). The description of the test cases follows. The first line of each test case contains one integer n (1≤n≤2⋅105 ). The second line of each test case contains 2n distinct integers b1,b2,…,b2n (1≤bi≤109 ), denoting the sequence b . It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 .    Output For each test case, output 2n+1 distinct integers, denoting the sequence a (1≤ai≤1018 ). If there are multiple possible sequences, you can output any of them. The sequence a should satisfy the given conditions, and it should be possible to obtain b after deleting one element from a and shuffling the remaining elements. Example InputCopy 4 1 9 2 2 8 6 1 4 3 99 2 86 33 14 77 2 1 6 3 2 OutputCopy 7 9 2 1 8 4 6 9 86 99 2 77 69 14 33 4 6 1 2 3
最新发布
03-11
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值