作者：ACShiryu

時(shí)間：2011-8-4

原題：http://poj.org/problem?id=2262

Goldbach's Conjecture

Time Limit:?1000MS		Memory Limit:?65536K
Total Submissions:?24877		Accepted:?9798

Description

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 題目大意就是輸入一個(gè)不小于6的合數(shù)，把它表示成兩個(gè)質(zhì)數(shù)的和，如果有多個(gè)，數(shù)出相差最大的一組 這題就是簡(jiǎn)單的枚舉+素?cái)?shù)判定，沒什么技巧 行開始時(shí)分解合數(shù)時(shí)到sqrt（n）時(shí)停止，WA了一次，應(yīng)該是n/2

參考代碼：

1 #include<iostream>
2 #include<cstdlib>
3 #include<cstdio>
4 #include<cstring>
5 #include<algorithm>
6 #include<cmath>
7 using namespace std;
8 bool isprime ( int k )
9 {
10 int t = sqrt ( k + 0.5 ) ;
11 for ( int i = 2 ; i <= t ; i ++ )
12 if ( k % i == 0 )
13 return false ;
14 return true ;
15 }
16 int main()
17 {
18 int n ;
19 while ( scanf ("%d", &n) , n )
20 {
21 int i ;
22 int t = n / 2 ;
23 for ( i = 3 ; i <= t ; i += 2 )
24 if ( isprime ( i ) && isprime ( n - i ) )
25 break ;
26 printf ( "%d = %d + %d\n" , n , i , n - i ) ;
27 }
28 return 0;
29 }

　　

轉(zhuǎn)載于:https://www.cnblogs.com/ACShiryu/archive/2011/08/04/2127966.html

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