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
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
Sample Input
8
20 42 0
Sample Output
8 = 3 + 5 20 = 3 + 17 42 = 5 + 37 这个题的意思就是说偶数可以由两个奇数而且还是素数的和组成，所以我们就求和呗
代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=1e6;
int data[maxn];
void sushu()
{
memset(data,1,sizeof(data));
for(int i=2;i<maxn;i++)
{
if(data[i]) for(long long j=(long long)i*i;j<maxn;j+=i)
data[j]=0;
}
}
bool odd(int m)
{
if(m%2)return 1;
return 0;
}
int main()
{
sushu();
int m;
while(cin>>m&&m)
{
cout<<m<<" = ";
for(int i=3;i<=m/2;i++)
{
int n=m-i;
if(odd(i)&&odd(n))
{
if(data[i]&&data[n])
{
cout<<i<<" + "<<n<<endl;
break;
}
}
}
}
return 0;
}
