POJ 2262 Goldbach's Conjecture

浏览数：21 / 时间：2015年06月12日

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 40455		Accepted: 15485

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

Sample Output

8 = 3 + 5
20 = 3 + 17
42 = 5 + 37

线性筛素数
CODE:

#include <iostream>
#include <cstdio>
#include <cstring>
#define REP(i, s, n) for(int i = s; i <= n; i ++)
#define REP_(i, s, n) for(int i = n; i >= s; i --)
#define MAX_N 1000000 + 10
#define MAX_M 80000

using namespace std;

int n, pri[MAX_M];
bool check[MAX_N];

int Prime(){
    memset(check, 0, sizeof(check));
    int tot = 0; check[1] = 1;
    REP(i, 2, MAX_N){
        if(!check[i]) pri[++ tot] = i;
        REP(j, 1, tot){
            if(i * pri[j] > MAX_N) break; check[i * pri[j]] = 1;
            if(i % pri[j] == 0) break;
        }
    }
    return tot;
}

int main(){
    int tot = Prime();
    while(scanf("%d", &n) != EOF){
        if(n == 0) break;
        REP(i, 1, tot){
            if(!check[n - pri[i]] && pri[i] % 2 != 0){
                printf("%d = %d + %d\n", n, pri[i], n - pri[i]);
                break;
            }
        }
    }
    return 0;
}