LightOJ1259 Goldbach`s Conjecture —— 素数表

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

1259 - Goldbach`s Conjecture

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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：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <cmath>
 7 #include <queue>
 8 #include <stack>
 9 #include <map>
10 #include <string>
11 #include <set>
12 using namespace std;
13 typedef long long LL;
14 const int INF = 2e9;
15 const LL LNF = 9e18;
16 const int MOD = 1e9+7;
17 const int MAXN = 1e7+10;
18 
19 bool notprime[MAXN];
20 int prime[1000010];
21 void getPrime()
22 {
23     memset(notprime, false, sizeof(notprime));
24     notprime[0] = notprime[1] = true;
25     prime[0] = 0;
26     for (int i = 2; i<=MAXN; i++)
27     {
28         if (!notprime[i])prime[++prime[0]] = i;
29         for (int j = 1; j<=prime[0 ]&& prime[j]<=MAXN/i; j++)
30         {
31             notprime[prime[j]*i] = true;
32             if (i%prime[j] == 0) break;
33         }
34     }
35 }
36 
37 int main()
38 {
39     getPrime();
40     int T, n, kase = 0;
41     scanf("%d",&T);
42     while(T--)
43     {
44         scanf("%d",&n);
45         int ans = 0;
46         for(int i = 1; prime[i]<=n/2; i++)
47             if(!notprime[n-prime[i]])
48                 ans++;
49         printf("Case %d: %d
", ++kase,ans);
50     }
51     return 0;
52 }

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