# hdu4715之素数罗选

www.MyException.Cn  网友分享于：2013-09-10  浏览：6次
hdu4715之素数筛选

# Difference Between Primes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 673    Accepted Submission(s): 216

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

```#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<iomanip>
#define INF 99999999
using namespace std;

const int MAX=1000000+10;
bool prime[MAX];
int s[200]={2},size;

void Prime(){//素数筛选
for(int i=2;2*i<MAX;++i)prime[2*i]=true;
for(int i=3;i*i<MAX;i+=2){
if(!prime[i]){
s[++size]=i;
for(int j=i*i;j<MAX;j+=i)prime[j]=true;
}
}
}

int main(){
Prime();
int t,n,m,i;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
m=n>0?n:-n;
for(i=0;i<=size;++i){
if(!prime[s[i]+m])break;
}
if(n>0)printf("%d %d\n",s[i]+m,s[i]);
else printf("%d %d\n",s[i],s[i]+m);
}
return 0;
}```