hdu 4715 Difference Between Primes （打表 枚举）

Difference Between Primes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 528    Accepted Submission(s): 150

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

思路：

打个素数表，然后枚举b，判断a就够了。

ps：汗，开始以为xat为正数，一直WA，原来题目说的是绝对值。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define maxn 10000005
using namespace std;

int n,m,cxx,ans;
bool vis[maxn];
int prime[664580];

void sieve(int nn)
{
    int i,j,mm;
    mm=int(sqrt(nn+0.5));
    memset(vis,0,sizeof(vis));
    for(i=2;i<=mm;i++)
    {
        if(!vis[i])
        {
            for(j=i*i;j<=nn;j+=i)
            {
                vis[j]=1;
            }
        }
    }
}
int get_prime(int nn)
{
    int i,c=0;
    sieve(nn);
    for(i=2;i<=nn;i++)
    {
        if(!vis[i]) prime[c++]=i;
    }
    return c;
}
bool isprime(int x)
{
    int i,j,t;
    t=sqrt(x+0.5);
    for(i=2;i<=t;i++)
    {
        if(x%i==0) return false ;
    }
    return true ;
}
int main()
{
    int i,j,t,a,b,flag,sgn;
    cxx=get_prime(10000000);  // 664579
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        flag=0;
        if(n==0)
        {
            printf("2 2\n");
            continue ;
        }
        if(n>0) sgn=1;
        else sgn=0,n=-n;
        for(i=0;i<cxx;i++)
        {
            b=prime[i];
            a=b+n;
            if(isprime(a))
            {
                flag=1;
                if(sgn) printf("%d %d\n",a,b);
                else printf("%d %d\n",b,a);
                break ;
            }
        }
        if(!flag) printf("FAIL\n");
    }
    return 0;
}