原题链接：
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
Note
An integer is said to be prime, if it is divisible by exactly two different integers. First few primes are 2, 3, 5, 7, 11, 13, …
题意：
输出t个测试样例，每一个都只含有一个n（4<=n<=107），找出n能否由两个素数相加得到，若是能找出有几种组合。输出这样的组合数。
题解：
数据范围很小，打一个素数表，依次判断两个加数是否都是素数即可。（注意不要有重复的：（2，3）（3，2）只算一种）
附上AC代码：

#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
using namespace std;
int prime[700000];
bool vis[10000005];
int cnt=0;
void el(int n)//素数打表
{
    int m=sqrt(n+0.5);
    memset(vis,0,sizeof(vis));
    for(int i=2;i<=m;i++)
    {
        if(!vis[i])
        for(int j=i*i;j<=n;j+=i)
        {
            vis[j]=1;
        }
    }
    for(int i=2;i<=n;i++)
    {
        if(!vis[i])
            prime[cnt++]=i;
    }
}
int main()
{
    el(10000000);
    int t,n;
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++)
    {
        int sum=0;
        scanf("%d",&n);
        for(int i=0;i<cnt;i++)
        {
            if(prime[i]>n-prime[i])//若是prime超过了n的一半，则所有的可能都已经找到了，剩下的都是重复的
                break;
            if(!vis[n-prime[i]])//判断是否为素数
            {
                sum++;
            }
        }
        printf("Case %d: %d\n",cas,sum);
    }
    return 0;
}

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