hdu4715 Difference Between Primes

Difference Between Primes

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

Total Submission(s): 0    Accepted Submission(s): 0


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水题,

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
using namespace std;
#define mod 1000000007

int ans;
int prime[1050000],pans,p[1050000];
int re;
int n;
int init()
{
int i;
memset(prime,0,sizeof(prime));
prime[0]=prime[1]=1;
for(i=2;i<=1000050;i++)
{
if(!prime[i])
for(int j=i+i;j<1000500;j=j+i)
{
prime[j]=1;
}
}
pans=0;
for(i=2;i<=1000000;i++)
{
if(!prime[i])
p[pans++]=i;
}
return 1;
}
int main()
{
init();
int tcase,i;
scanf("%d",&tcase);
bool flag=true;
while(tcase--)
{
scanf("%d",&n);
flag=true;
for(i=0;i<pans;i++)
{
if(p[i]>=n&&prime[p[i]-n]==0)
{
printf("%d %d\n",p[i],p[i]-n);
flag=false;
break;
}
}
if(flag)
printf("FAIL\n");
}
return 0;
}

 

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