F

F - Goldbach`s Conjecture LightOJ - 1259

题目链接：https://vjudge.net/contest/70017#problem/F

题目：

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, ...

题意：给你一个偶数，算出多少种a+b=n的可能，其中a,b要是素数且a<b思路;素数筛，打表，

//
// Created by hanyu on 2019/8/12.
//
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<cmath>
#include<string>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<iostream>
#include<algorithm>
typedef long long ll;
const int maxn=1e7+5;
using namespace std;
bool isprime[maxn];
int prime[maxn/10],pos;
void getp()
{
    pos=0;
    memset(isprime,false,sizeof(isprime));
    isprime[0]=isprime[1]=true;
    for(long long i=2;i<maxn;i++){
        if(!isprime[i])
        {
            prime[pos++]=i;
            for(long long j=i*i;j<maxn;j+=i){
                isprime[j]=true;
            }
        }
    }
}

int main()
{
    getp();
    int T,n,kase=1,ans;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        ans=0;
        for(int i=0;prime[i]<=n/2;i++)
        {
            if(isprime[n-prime[i]]==false)
                ans++;
        }
        printf("Case %d: %d
",kase++,ans);
    }
    return 0;
}

注意数据范围，我re了很多次。。。。