poj 2262 Goldbach's Conjecture 筛法 哥德巴赫猜想

poj 2262 Goldbach's Conjecture 筛法 哥德巴赫猜想

Goldbach's Conjecture

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 43657 Accepted: 16675

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

题意:给一个数n,写成最小的两个素数的和

思路:从2开始找素数m,当n-m也是素数时说明找到答案,由哥德巴赫猜想知这两个数一定是存在的;

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<queue>using namespace std;#define M 1000000bool visit[M];int p[M];void iii(){    int np=0;    memset(visit,true,sizeof(visit));    visit[1]=0;    for(long long  i=2; i<M; ++i)    {        if(visit[i]==true)        {            p[np++]=i;            for(long long  j=i*i; j<M; j+=i)            {                visit[j]=false;            }        }    }}int main(){    iii();    int i,n,m,d;    while(~scanf("%d",&n))    {        if(n==0) break;        m=n/2;        for(i=0;p[i]<=m;i++)        {            d=n-p[i];            if(visit[d]==1)            {                printf("%d = %d + %d\n",n,p[i],d);break;            }        }    }     return 0;}