K - Goldbach's Conjecture解题报告

K - Goldbach's Conjecture

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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

这题超时了N遍，就是不记得筛选法求素数了，哎~~~这次可要牢记了，其他的就不多说了。。。

#include<iostream>#include<time.h>short prime[1000000];using namespace std;int main(){int i,j,s=0,n,num;memset(prime,0,sizeof(prime));for(i=2;i<1000000;i++)for(j=2;i*j<1000000;j++){if(prime[i*j]==0)prime[i*j]=1;}while(scanf("%d",&n)&&n!=0){for(i=3;i*2<=n;i+=2)if(prime[i]==0&&prime[n-i]==0){printf("%d = %d + %d\n",n,i,n-i);goto end;}printf("Goldbach's conjecture is wrong.\n");end:;}return 0;}