hduoj 4715 Difference Between Primes 2013 ACM/ICPC Asia Regional Online —— Warmup

http://acm.hdu.edu.cn/showproblem.php?pid=4715

Difference Between Primes

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

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 x at the next n lines. The absolute value of x is not greater than 10^6.

 

Output

For each number x tested, outputs two primes a and b at one line separated with 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

分析：

这道题就是求一个整数用两个素数（大于等于2）的差表示出来，要求两个素数在满足条件的情况下最小。

AC代码：

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <vector>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#pragma warning(disable:4786)

using namespace std;

const int INF = 0x3f3f3f3f;
const int MAX = 1000000 + 10;
const double eps = 1e-8;
const double PI = acos(-1.0);

int a[MAX];

int main()
{
    int i , j ;
    memset(a , 0 , sizeof(a));
    a[1] = 1; a[0] = 1;
    for(i = 2;i <= sqrt(MAX);i++)
    {
        if(a[i] == 0)
            for(j = i * i;j <= MAX;j += i)
                a[j] = 1;
    }
    int n,m;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&m);
        int t = abs(m);
        for(i = t;i < MAX;i++)
            if(!a[i] && !a[i - t])
            {
                if(m > 0)
                {
                    printf("%d %d
",i,i - t);
                    break;
                }
                else
                {
                    printf("%d %d
",i - t,i);
                    break;
                }
            }
        if(i == MAX)
            printf("FAIL
");
    }
    return 0;
}