HOJ1459 Goldbach's Conjecture

Goldbach's Conjecture

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	Source : University of Ulm Internal Contest 1998
	Time limit : 1 sec		Memory limit : 32 M

Submitted : 2950, Accepted : 1033

In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:

Every even number greater than 4 can be

written as the sum of two odd prime numbers. 

For example: 

8 = 3 + 5. Both 3 and 5 are odd prime numbers. 
20 = 3 + 17 = 7 + 13. 
42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23. 

Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.) 

Anyway, your task is now to verify Goldbach's conjecture for all even numbers less than a million.

Input

The input will contain one or more test cases. 

Each test case consists of one even integer n with 6 <= n < 1000000. 

Input will be terminated by a value of 0 for n.

Output

For each test case, print one line of the form n = a + b, where a and b are odd primes. Numbers and operators should be separated by exactly one blank like in the sample output below. If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized. If there is no such pair, print a line saying "Goldbach's conjecture is wrong."

Sample Input

8
20
42
0

Sample Output

8 = 3 + 5
20 = 3 + 17
42 = 5 + 37

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/*
  题意：哥德巴赫猜想，大于四的偶数可以分解为两个奇素数之和
  思路：直接枚举所有的，看看是否符合条件，从小到大枚举到中间的话符合条件的也就是差值最大的！
*/
#include 
#include
#include
using namespace std;
int prime(int x)
{
    double M;
    int i;
    M=sqrt(x);
    for(i=2; i<=M; i++)
        if(x%i==0)
            return 0;
    return 1;
}
int main()
{
    int n,i;
    while(cin >> n&&n)
    {
        for(i=2; i