文章标题

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, …

解题报告

先得对$10^7$ 内的素数打表只有664580个，如此数据量级就减小了100倍

判断素数a+b=n，只要取素数a，在判断 n-a 是否为素数即可复杂度o(φ(n))

#include<stdio.h>
#include<algorithm>
#define MAX_N 10000020
#define P_N 664580
using namespace std;
bool vis[MAX_N];
int p[P_N];

void init(){
    int cnt=0;
    for(int i=2;i*i<=MAX_N;i++)
        if(!vis[i])
            for(int k=i*i;k<MAX_N;k+=i)
                vis[k]=true;
    for(int i=2;i<MAX_N;i++)
        if(!vis[i])
            p[cnt++]=i;
}

int main()
{
    init();int T;
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        int n;
        scanf("%d",&n);
        int ans=0;
        for(int i=0;p[i]<=n/2;i++){
            if(!vis[n-p[i]]) ans++;
        }
        printf("Case %d: %d\n",t,ans);
    }
    return 0;
}