poj 2262 Goldbach's Conjecture

浏览数：19 / 时间：2015年06月12日

Goldbach‘s Conjecture

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 39353		Accepted: 15077

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

Source

Ulm Local 1998

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string>
 5 #include<cstring>
 6 #include<vector>
 7 using namespace std;
 8 #define max 1000000
 9 int prime[max+5];
10 bool vis[max+5];
11 void get_prime(int n){
12     memset(vis,false,sizeof(vis));
13     vis[0]=vis[1]=true;
14     int i=2,index=0;
15     for(;i<=n;i++){
16         if(!vis[i]){
17             prime[index++]=i;
18             //vis[i]=false;
19         }
20         int j=0;
21         for(;j<index&&prime[j]*i<=n;j++){//????
22             vis[prime[j]*i]=true;
23             if(!(i%prime[j]))
24             break;
25         }
26     }
27 }
28 int main(){
29     int n;
30     get_prime(max);
31     while(cin>>n&&n){
32         int i=3,half=n/2;
33         for(;i<=half;i+=2){//等号一定要取到，存在数等于两个相同质数之和的
34             if(!vis[i]&&!vis[n-i]){
35                 break;
36             }
37         }
38         if(i>half)
39         cout<<"Goldbach‘s conjecture is wrong."<<endl;
40         else
41         cout<<n<<" = "<<i<<" + "<<n-i<<endl;
42     }
43     return 0;
44 }