POJ 2262 Goldbach's Conjecture

最新推荐文章于 2021-02-02 10:09:10 发布

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最新推荐文章于 2021-02-02 10:09:10 发布

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原文链接：http://www.cnblogs.com/jun930123/archive/2012/08/16/2642315.html

本文验证了哥德巴赫提出的猜想，即每一大于4的偶数都能表示为两个奇素数之和。通过算法筛选奇素数，并在输入的偶数范围内找到符合条件的奇素数对。实例展示了该猜想在特定数值下的应用。

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I - Goldbach's Conjecture

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2262

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

Sample Output

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

View Code

 1 #include<stdio.h>
 2 int prime[101000],f[1010000],count=0 ;
 3 void getPrime(){
 4     for(int i=2;i<=1000000;i++)
 5         if(!f[i]){
 6             prime[count++] = i;
 7             for(__int64 j=(__int64)i*i;j<=1000000;j+=i)
 8                 f[j] = 1;
 9         }
10 }
11 int main(){
12     getPrime();
13     int n;
14     while(scanf("%d",&n) , n){
15         for(int i=0;i<count;i++)
16             if(f[n-prime[i]] == 0){
17                 printf("%d = %d + %d\n",n,prime[i],n-prime[i]) ;
18                 break;
19             }
20     }
21     return 0;
22 }