hdu 4715 Difference Between Primes （二分查找）

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

 

        题意：找两个素数相差为N的素数。

        先个所有素数打表。然后对每个N进行二分查找。

        注意lower_bound二分查找的用法。

      

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int zj[5000000];
int ssb[5000000],pn;
void f()
{
    int i,j;
    memset(zj,0,sizeof(zj));
    zj[0]=zj[1]=1;
    pn=0;
    for (i=2;i<=5000000;i++)
    {
        if (!zj[i]) {ssb[pn]=i;pn++;}
        for (j=0;j<pn;j++)
        {
            if (i*ssb[j]>5000000) break;
            zj[i*ssb[j]]=1;
            if (i%ssb[j]==0) break;
        }
    }
}
int main()
{
    int t,n,i,c;
    f();
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d",&n);
        for (i=0;i<pn;i++)
        {
           c=lower_bound(ssb,ssb+pn,ssb[i]+n)-ssb;
           if (ssb[c]==ssb[i]+n)
           {
               printf("%d %d\n",ssb[c],ssb[i]);
               break;
           }
        }
    }
}