poj 2262 Goldbach's Conjecture(素数筛选法)

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

Goldbach's Conjecture

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 34323		Accepted: 13169

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

Source

Ulm Local 1998

 

【题解】：

题意：输入一个偶数n，将n分成两个素数和，找到第一组就可以（n=a+b，a尽量小）

n最大1000000

方法：素数筛选法

【code】：

/**
Judge Status:Accepted    Memory:4852K
Time:344MS    Language:G++
Code Lenght:797B    Author:cj
*/
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

#define N 1000000
using namespace std;

int prim[N+10];
int arrPrim[N+10];

int main()
{
    int i,j;
    memset(prim,0,sizeof(prim));
    for(i=2;i*i<=N;i++)  //素数筛选法
    {
        if(!prim[i])
        {
            for(j=2;j*i<=N;j++)
            {
                prim[j*i]=1;
            }
        }
    }
    int arr_cnt=0;
    for(i=3;i<=N;i++)
    {
        if(!prim[i]) arrPrim[arr_cnt++]=i;  //一次遍历赛选出来的素数存入数组，遍历用
    }
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(i=0;i<arr_cnt;i++)
        {
            if(!prim[n-arrPrim[i]])
            {
                printf("%d = %d + %d\n",n,arrPrim[i],n-arrPrim[i]);
                break;
            }
        }
    }
    return 0;
}