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, ...
题意：给你一个n，找出有多少组 两个素数的和等于n；
直接从1到n/2遍历的话会超时。所以遍历所有的素数找n-p[i]是否为素数。
#include<map>
#include<stack>
#include<queue>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define maxn 10000001
#define maxm 1000000000005
#define mod 1000000007
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
bool vis[maxn];
int p[1000000];
int tot;
void getp(){
tot=0;
mem(vis,true);
vis[1]=vis[0]=false;
for(ll i=2;i<=maxn;i++){
if(vis[i]){
p[tot++]=i;
for(ll j=i*i;j<=maxn;j+=i)vis[j]=false;
}
}
}
int main(){
getp();
int t,test=0;scanf("%d",&t);
while(t--){
int x;scanf("%d",&x);
int ans=0;
for(int i=0;i<tot&&p[i]<=x/2;i++){
if(vis[x-p[i]])
ans++;
}
printf("Case %d: %d
",++test,ans);
}
}