20140911 【 初等数论 】 poj 2262 Goldbach's Conjecture

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Goldbach's Conjecture

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 37916 Accepted: 14588

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

820420

Sample Output

8 = 3 + 520 = 3 + 1742 = 5 + 37

Source

Ulm Local 1998

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这题没什么好讲的，

只是为了记录一下压缩形式的素数筛选。

毕竟在C++中 bool 类型也是占一个字节的。

挺浪费空间。

改用位标记，可以压缩8倍空间。。

当然，时间上还是没有优化。。

（至于额外的时间花费？？  要相信位运算的力量）

#include <iostream>#include <algorithm>using namespace std;#define MAXN 1000000char prime[MAXN/7];int p[MAXN+10];int np, n;void init(){    prime[0] = 3;    np = 0;    for(int i=2; i<MAXN; i++)        if( !( prime[i>>3]&(1<<(i&7)) ) ){            p[np++] = i;            for(int j=i*i; j<MAXN; j+=i)                prime[j>>3] |= 1<<(j&7);        }}int main(){    init();    while( cin>>n && n ){        int mid = np>>1|1, i, f;        for(i=0; i<mid; i++){            f = upper_bound(p, p+np, n-p[i]) - p-1;            if( f>=0 && p[i]+p[f]==n )    break;        }        if( i!=mid )    cout<<n<<" = "<<p[i]<<" + "<<p[f]<<endl;        else            cout<<"Goldbach's conjecture is wrong."<<endl;    }    return 0;}