Goldbach`s Conjecture（数论）素数筛

题目：https://vjudge.net/contest/242286#problem/A

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, ...

分析：找到两个素数和为n，只需要找一个比n/ 2小的，一个大的，我们可以实现大个表，存下[1,10000007]所有素数，之后在里边找；不过用欧氏筛需要做一步优化，在找B的时候需要改一下，因为prime[]里边存的就是素数，所以限制条件可以用它，时间会缩短。

#include <iostream>
#include <stdio.h>
#include <string.h>
#define MAXN 10000005
using namespace std;
int N;
bool isprime[MAXN+1];
int prime[MAXN/10], tot; //防止MLE
void getPrime() {
    memset(isprime, true, sizeof(isprime));
    isprime[0] = isprime[1] = false;
    tot = 0;
    for(int i = 2; i < MAXN; i++) {
        if(isprime[i]) prime[tot++] = i;
        for(int j = 0; j < tot && prime[j] <= MAXN / i; j++) {
            isprime[i * prime[j]]=false;
            if(i % prime[j] == 0) break;
        }
    }
}


int main()
{
    getPrime();
    int t;
    int Case=0;
    scanf("%d",&t);
    while(t--)
    {
        int cnt=0;
        int b;
        scanf("%d",&N);

        for(int i=0;prime[i]<=N/2;i++)//用欧拉筛，需要优化这里。
        {
                 b=N-prime[i];
            if(isprime[b])cnt++;
        }
        printf("Case %d: %d\n",++Case,cnt);
    }
    return 0;
}