【原】 POJ 2262 Goldbach's Conjecture 筛素数 解题报告

 

http://poj.org/problem?id=2262

 

方法：
首先筛得1000000以下的素数表，复杂度N*lglgN
再用测试值减去素数表中从小到大的值，判断如果该值是素数，则不需要再接着找了，直接输出。复杂度N

注意点：
1、由于N很大，所以筛素数在计算i*i时可能会超过int的范围，所以需要采用__int64
2、需要符合要求的a,b使得a+b=c的方法：
       a)枚举的方法。复杂度n^2
       b)顺序扫描符合要求的元素a，判断c-a是否也符合要求。复杂度n

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

 

#include <iostream>

using namespace std ;

const __int64 N = 1000001 ;
const __int64 M = 78500;
char table[N] = {0} ;
//int prime[M] ;

//0为素数
void GetPrimeTable()
{
    __int64 i ;
    __int64 j ;

    table[0]=table[1]=1;
    i=2 ;
    while( i<N )
    {
        for( j=i*i ; j<N ; j+=i )
            table[j]=1 ;
        do ++i ;
        while( i<N && table[i]==1 );
    }
    /*
    for( i=2,j=0 ; i<N ; ++i )
    {
        if( table[i]==0 )
            prime[j++]=i ;
    }
    */
}


void run2262()
{
    GetPrimeTable() ;

    int val ;
    int a,b ;
    int i,j ;
    int sum ;
    bool flag ;
    while(cin>>val && val!=0)
    {
        /* TLE
        //枚举 n^2
        flag = false ;
        for( i=0 ; i<M && flag==false ; ++i )
        {
            for( j=i+1 ; j<M ; ++j )
            {
                sum = prime[i]+prime[j] ;
                if( sum > val )
                {
                    flag = true ;
                    break ;
                }
                else if( sum == val )
                {
                    a = prime[i] ;
                    b = prime[j] ;
                    flag = true ;
                    break ;
                }
            }
        }
        */
        for( i=2 ; i<N ; ++i )
        {
            if( table[i]==0 && table[val-i]==0 )
            {
                a = i ;
                b = val-i ;
                break ;
            }
        }
        cout<<val<<" = "<<a<<" + "<<b<<endl ;
    }
}