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hdu 4715 Difference Between Primes (二分查找)

时间:2015-08-15 16:25:19      阅读:99      评论:0      收藏:0      [点我收藏+]

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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 }

 

hdu 4715 Difference Between Primes (二分查找)

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原文地址:http://www.cnblogs.com/pblr/p/4732532.html

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