HDU 4715 (素数筛选）（2015排位1-H）

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

3 6 10 20

 

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;
}