Goldbachs Conjecture （素数打表）

Description

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：找出两个素数相加等于n的对数，将素数用一个数组记录下来，元素数量一定小于<10000000/2，防T。

代码如下：

#include <iostream>
#include <queue>
#include <math.h>
#include <stdio.h>
#define PI acos(-1)
using namespace std;
#define MAXN 10000010
#define LL long long
bool v[MAXN]={0};
int prime[666666],cnt;
void db()
{
    cnt=0;
    v[1]=1;
    for(int i=2; i<=10000000; i++)
    {
        if(!v[i])
        {
            prime[cnt++]=i;//记录素数的数组
            for(int j=i+i; j<=10000000; j+=i)
                v[j]=1;
        }
    }
}
int main()
{
db();
    int t,cc=0;
    cin>>t;
    while(t--)
    {
    int n;
        cin>>n;
        int ans=0;
        for(int i=0; i<cnt; i++)
        {
            if(prime[i]>=n/2+1)
                break;
            if(!v[n-prime[i]]&&n>=prime[i]*2)//确保双素数和条件3
                ans++;
        }
        cout<<"Case "<<++cc<<": "<<ans<<endl;
    }
    return 0;
}