F - Goldbach`s Conjecture (哥德巴赫猜想）
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
1. 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, ...
题解：暴力肯定T，一个素数表也T，再加个vis就A了。
#include
#include
#include
using namespace std;
const int N=10000000;
int isprime[700000]; //从前往后存素数值
bool vis[N]; //判断素数
int tot;
void prime()
{
memset(vis,1,sizeof(vis)); //初始化为1
vis[1]=0;
for(int i=2;i=N) //边缘条件，一定要加，不然会RE
break;
vis[i*isprime[j]]=0; //该数一定有因子，取消标记
//if(i%isprime[j]==0) //这两行剪枝，可不要
// break;
}
}
}
int main()
{
int t,k=1;
scanf("%d",&t);
prime();
while(t--)
{
int x;
int sum=0;
scanf("%d",&x);
for(int i=0; isprime[i]<=x/2; i++)
{
if(vis[x-isprime[i]]) //从头开始判断素数，isprime[i]一定是素数，只需判断x-isprime是否为素数即可
sum++;
}
cout<<"Case "<