题目：
Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states:
Every even integer, greater than 2, can be expressed as the sum of two primes [1].
Now your task is to check whether this conjecture holds for integers up to 107.
Input
Input starts with an integer T (≤ 300), denoting the number of test cases.
Each case starts with a line containing an integer n (4 ≤ n ≤ 107, n is even).
Output
For each case, print the case number and the number of ways you can express n as sum of two primes. To be more specific, we want to find the number of (a, b) where
1) Both a and b are prime
2) a + b = n
3) a ≤ b
Sample Input
2
6
4
Sample Output
Case 1: 1
Case 2: 1
题意：
判断一个数是否能由两个素数相加得到，如果可以，写出有几组；
分析：
直接素数打表，然后判断n减去这个素数是否为素数即可；需要注意的是，素数打表的时候，我使用j=i*i的发现一直wr，改为j=2*i可以过，而且在记录素数时最好使用bool,否则会超限；
代码：
#include<stdio.h>
#include<math.h>
using namespace std;
#define maxn 10000010
bool p[maxn];
int prim[666666];
int t,k=0;
int n;
void primary()
{
p[1]=1;
for(int i=2;i<=maxn;i++)
{
if(!p[i])
{
prim[k++]=i;
for(int j=i+i;j<=maxn;j+=i)
{
p[j]=1;
}
}
}
return ;
}
int main()
{
int casee=0;
scanf("%d",&t);
primary();
while(t--)
{
scanf("%d",&n);
int ans=0;
for(int i=0;i<k;i++)
{
if(prim[i]>=n/2+1)//因为是从小到大，因此只要这个素数大于n/2那就说明后面的不用判断了
break;
if(p[n-prim[i]]==0)
ans++;
}
printf("Case %d: %d\n",++casee,ans);
}
return 0;
}