HDU 4715 (素数筛选）（2015排位1-H）

最新推荐文章于 2022-12-20 10:26:01 发布

Rabbee

最新推荐文章于 2022-12-20 10:26:01 发布

阅读量702

点赞数

CC 4.0 BY-SA版权

分类专栏：素数判定文章标签：素数

本文链接：https://blog.youkuaiyun.com/Rabbee/article/details/44598165

素数判定专栏收录该内容

1 篇文章

订阅专栏

这篇博客详细介绍了HDU 4715竞赛题目，内容涉及素数筛选的方法和解决方案，适合算法爱好者和竞赛编程初学者学习。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4715

Description

All you know Goldbach conjecture.That is to say, Every even integer greater than 2 can be expressed as the sum of two primes. Today, skywind present a new conjecture: every even integer can be expressed as the difference of two primes. To validate this conjecture, you are asked to write a program.

Input

The first line of input is a number nidentified the count of test cases(n<10^5). There is a even number xat the next nlines. The absolute value of xis not greater than 10^6.

Output

For each number xtested, outputstwo primes aand bat one line separatedwith one space where a-b=x. If more than one group can meet it, output the minimum group. If no primes can satisfy it, output 'FAIL'.

Sample Input

Sample Output

  
    11 5
13 3
23 3 

-----------------------------------------------------------

拿到题就打开书，抄了素筛的模板上去，没有另起数组保存素数，判定时直接套两个循环，虽然没有超时，但是WA了。
后来发现，我考虑的负数的情况，用了两个负数表示才错了。还是做题不够，对素数也不熟悉。
-----------------------------------------------------------


   
    #include
     
      
#include
      
       
#include
       
        
#include
        
         
#include
         
          
using namespace std;
typedef long long ll;

bool is_prime[1000010];
int main()
{
    memset(is_prime, true, sizeof(is_prime));
    is_prime[0] = is_prime[1] = false;
    for(int i = 2; i <= 1000010; i++) {
        if(is_prime) {
            for(int j = 2*i; j <= 1000010; j += i) is_prime[j] = false;
        }
    }
    int T;
    scanf("%d", &T);
    while( T-- )
    {
        int n;
        scanf("%d", &n);
        int j = -1, k = -1;
        if(n >= 0){
            for(int i = 1; i <= 1000010 - n - 1; i++)
            {
                if(is_prime[n+i] && is_prime[i])
                {
                    j = n+i;
                    k = i;
                    break;
                }
            }
            if(j != -1) printf("%d %d\n", j, k);
            else puts("FAIL");
        }
        else{
                n = -n;
            for(int i = 1; i <= 1000010 - n - 1 ; i++)
            {
                if(is_prime[n+i] && is_prime[i])
                {
                    j = n+i;
                    k = i;
                    break;
                }
            }
            if(j != -1) printf("%d %d\n", k, j);
            else puts("FAIL");
        }
    }
    return 0;
}