HDU 4715 Difference Between Primes

Difference Between Primes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 338    Accepted Submission(s): 78

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

 

Recommend

liuyiding

 

 

打出素数表，

枚举小的，

 

然后二分查找大的

/* *******************************************
Author       : kuangbin
Created Time : 2013年09月08日 星期日 13时30分10秒
File Name    : 1010.cpp
******************************************* */

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

const int MAXN = 5000000;
int prime[MAXN];
void getPrime()
{
    memset(prime,0,sizeof(prime));
    for(int i = 2;i <= MAXN;i++)
    {
        if(!prime[i])prime[++prime[0]] = i;
        for(int j = 1;j <= prime[0] && prime[j] <= MAXN/i;j++)
        {
            prime[prime[j]*i] = 1;
            if(i % prime[j] == 0)break;
        }
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    getPrime();
    int T;
    int n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i = 1;i < prime[0];i++)
        {
            int t = lower_bound(prime+1,prime + prime[0],prime[i]+n)- prime;
            if(prime[t] == prime[i] + n)
            {
                printf("%d %d
",prime[t],prime[i]);
                break;
            }
        }
    }
    return 0;
}