题目链接：https://vjudge.net/problem/LightOJ-1259

1259 - Goldbach`s Conjecture

Time Limit: 2 second(s)

Memory Limit: 32 MB

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

Output for Sample Input

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：

哥德巴赫猜想：任何一个大于2的偶数，都可以是两个素数的和。给出一个偶数，判断有多少对素数的和是这个数。

题解：

由于n<=1e7，所以我们可以先筛选出1e7范围内的素数，然后再枚举每一个素数进行判断。

代码如下：

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <vector>

 #include <cmath>

 #include <queue>

 #include <stack>

 #include <map>

 #include <string>

 #include <set>

 using namespace std;

 typedef long long LL;

 const int INF = 2e9;

 const LL LNF = 9e18;

 const int MOD = 1e9+;

 const int MAXN = 1e7+;

 bool notprime[MAXN];

 int prime[];

 void getPrime()

 {

     memset(notprime, false, sizeof(notprime));

     notprime[] = notprime[] = true;

     prime[] = ;

     for (int i = ; i<=MAXN; i++)

     {

         if (!notprime[i])prime[++prime[]] = i;

         for (int j = ; j<=prime[ ]&& prime[j]<=MAXN/i; j++)

         {

             notprime[prime[j]*i] = true;

             if (i%prime[j] == ) break;

         }

     }

 }

 int main()

 {

     getPrime();

     int T, n, kase = ;

     scanf("%d",&T);

     while(T--)

     {

         scanf("%d",&n);

         int ans = ;

         for(int i = ; prime[i]<=n/; i++)

             if(!notprime[n-prime[i]])

                 ans++;

         printf("Case %d: %d\n", ++kase,ans);

     }

     return ;

 }

巴特西

LightOJ1259 Goldbach`s Conjecture —— 素数表

Input

Output

Sample Input

Output for Sample Input

Note

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