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
Sample Output
Source
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;
}