F - Goldbach`s Conjecture kuangbin 基础数论

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

Hint

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, ...

2. #include<iostream>
   #include<cstdio>
   #include<cmath>
   #include<cstring>
   #include<sstream>
   #include<algorithm>
   #include<queue>
   #include<deque>
   #include<iomanip>
   #include<vector>
   #include<cmath>
   #include<map>
   #include<stack>
   #include<set>
   #include<fstream>
   #include<memory>
   #include<list>
   #include<string>
   using namespace std;
   typedef long long LL;
   typedef unsigned long long ULL;
   #define MAXN 10000001
   #define L 31
   #define INF 1000000009
   #define eps 0.00000001
   /*
   打表 把所有素数存到一个vector中
   然后用一个map保存所有和出现的次数
   然后直接找就可以
   */
   bool notprime[MAXN];
   vector<int> prime;
   
   void Init()
   {
       memset(notprime, false, sizeof(notprime));
       notprime[1] = true;
       for (int i = 2; i < MAXN; i++)
       {
           if (!notprime[i])
           {
               prime.push_back(i);
               for (int j = i + i; j < MAXN; j += i)
                   notprime[j] = true;
           }
       }
   }
   int main()
   {
       Init();
       int T,n;
       cin >> T;
       for(int cas=1;cas<=T;cas++)
       {
           cin >> n;
           vector<int>::iterator p = lower_bound(prime.begin(), prime.end(), n/2);
           //cout << *p << endl;
           int cnt = 0;
           for (vector<int>::iterator it = prime.begin(); it <= p && *it<=n/2; it++)
           {
               if (!notprime[n - *it])
               {
                   //cout << *it << ‘ ‘ << n - *it << endl;
                   cnt++;
               }
           }
           printf("Case %d: %d\n", cas, cnt);
       }
       return 0;
   }