Goldbach`s Conjecture 素数筛选

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, ...

题意：

一个偶数=两个素数相加，给定偶数N,求满足要求的素数对

注：数组开1e7会MT,提前计算素数个数，1e7以内只有六万多，轻松过

#include<cstdio>
#include<cstring>
using namespace std;
const int N=10000005;
int tot=0,ans[666666];//数组开小一点
bool v[N];//bool必须开N

void prime()
{
    memset(v,true,sizeof(v));v[1]=false;
    for(int i=2;i<N;++i)
    {
        if(v[i]) ans[tot++]=i;
         for(int j=0;(j<tot)&&(i*ans[j]<N);++j)
         {
             v[i*ans[j]]=false;
             if(i%ans[j]==0) break;
         }
    }
}

int main()
{
    int t,n;
    scanf("%d",&t);
    prime();
    for(int c=1;c<=t;++c)
    {
        int cnt=0;
        scanf("%d",&n);
        for( int i=0;ans[i]<=n/2;++i)
            if(v[n-ans[i]]) cnt++;
        printf("Case %d: %d\n",c,cnt);
    }
    return 0;
}