hdu4715

Difference Between Primes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 970    Accepted Submission(s): 315

Problem 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

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

#include<iostream>
#include<stdio.h>
#include<string.h>
#define M 1000000
using namespace std;
int vit[1000001];
int flg[1000000];
int main()
{
    memset(flg,0,sizeof(flg));
    vit[1]=0;
    vit[2]=1;
    for(int i=3; i<=M; i++)//素数打表
        if(i%2)
            vit[i]=1;
        else
            vit[i]=0;
    for(int i=3; i<=M; i++)
        if(vit[i])
            for(int j=i*2; j<=M; j+=i)
                vit[j]=0;
    int cnt=0;
    for(int i=2; i<=M; i++)
        if(vit[i])
        {
            flg[++cnt]=i;
        }
    int t,n,flag,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        if(n==0)
        {
            printf("2 2
");
            continue;
        }
        flag=0;
        for(int i=1; i<=cnt; i++)
        {
            if(vit[n+flg[i]])
            {
                printf("%d %d
",n+flg[i],flg[i]);
                flag=1;
                break;
            }
        }
        if(flag==0)
            printf("FAIL
");
    }
    return 0;
}