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。
1 #include<cstdio>
2 #include<cstring>
3 #include<iostream>
4 #include<algorithm>
5 using namespace std;
6 int zj[5000000];
7 int ssb[5000000],pn;
8 void f()
9 {
10 int i,j;
11 memset(zj,0,sizeof(zj));
12 zj[0]=zj[1]=1;
13 pn=0;
14 for (i=2;i<=5000000;i++)
15 {
16 if (!zj[i]) {ssb[pn]=i;pn++;}
17 for (j=0;j<pn;j++)
18 {
19 if (i*ssb[j]>5000000) break;
20 zj[i*ssb[j]]=1;
21 if (i%ssb[j]==0) break;
22 }
23 }
24 }
25 int main()
26 {
27 int t,n,i,c;
28 f();
29 scanf("%d",&t);
30 while (t--)
31 {
32 scanf("%d",&n);
33 for (i=0;i<pn;i++)
34 {
35 c=lower_bound(ssb,ssb+pn,ssb[i]+n)-ssb;
36 if (ssb[c]==ssb[i]+n)
37 {
38 printf("%d %d\n",ssb[c],ssb[i]);
39 break;
40 }
41 }
42 }
43 }