素数筛 LightOJ

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

题意就是找一个偶数可以分解为几对素数之和，数据有些大，需要用筛法求素数。

#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stdio.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=1e7;
bool vis[maxn];
int pri[maxn/10];
void prime()
{
    memset(vis,0,sizeof(vis));
    int num=0;
    for(int i=2;i<maxn;i++)
    {
        if(!vis[i])       //原理就是每次标记所有约数有i的数，然后遍历所有没有被标记过的数
        {
            pri[++num]=i;
            for(int j=i*2;j<=maxn;j+=i)
            vis[j]=1;
        }
    }
}
int main()
{
    prime();
    int t,n;
    cin>>t;
    for(int cas=1;cas<=t;cas++)
    {
        cin>>n;
        int ans=0,tmp=n/2;
        for(int i=1;pri[i]<=tmp;i++)
        {
            if(vis[n-pri[i]]==0)
            {
                ans++;
            }
        }
        cout<<"Case "<<cas<<": "<<ans<<endl;
    }
    return 0;
}