POJ 2262 Goldbach's Conjecture(素数表分解质数)

Goldbach's Conjecture

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 41314 Accepted: 15827

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将一个数分解为两个素数相加，且差值最大的那一个，因为数据范围为100w，所以直接打一个素数表就行了。ac代码：
#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#define MAXN 1000100#define INF 0xfffffff#define MIN(a,b) a>b?b:ausing namespace std;int v[MAXN];void db(){int i,j;memset(v,0,sizeof(v));for(i=2;i<=1000000;i++){if(v[i]==0){for(j=i*2;j<=1000000;j+=i)v[j]=1;}}}int main(){db();int num;int i;while(scanf("%d",&num)&&num){int bz=0;for(i=2;i<num;i++){if(!v[i]&&!v[num-i]){printf("%d = %d + %d\n",num,i,num-i);bz=1;break;}}if(bz==0)printf("Goldbach's conjecture is wrong.\n");}return 0;}