POJ 2262 Goldbach's Conjecture 素数筛选

原题： http://poj.org/problem?id=2262

题目：

Goldbach’s Conjecture
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 41068 Accepted: 15725
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

思路：

对于给定一个数，找到两个奇素数，使他们的和为给定的数，如果找不到就输出：”Goldbach’s conjecture is wrong.”
对于多种答案，只需要求相差最大的。

这道题我们先用素数筛选，然后枚举它范围内的每个值和剩下的部分，这两个数是否都为奇素数。

代码：

#include <iostream>#include"stdio.h"#include"math.h"#include"string.h"#include"algorithm"using namespace std;typedef long long int lint;bool prime[1000005];void isprime(){    memset(prime,true,sizeof(prime));    for(int i=2;i<=1000000;i++)    {        if(prime[i])            for(int j=2*i;j<=1000000;j=j+i)            {                prime[j]=false;            }    }    prime[2]=false;}int main(){    //freopen("in.txt","r",stdin);    isprime();    int t;    while(scanf("%d",&t)!=EOF,t)    {        int i;        for(i=3;i<t;i++)        {            if(prime[i]&&prime[t-i]&&i<=t-i)                break;        }        if(i==t)  printf("Goldbach's conjecture is wrong.\n" );        else printf("%d = %d + %d\n",t,i,t-i);    }    return 0;}