Goldbach`s Conjecture（素筛水题）题解

Goldbach`s Conjecture

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

思路：

水题一只，叫你求n=a+b且a，b都是素数这样的ab的对数。只要素筛一下，就能做了。最好先储存一下10^7以内的素数，数答案时只要到n/2就行了（因为a<=b），不知道不弄会不会超时，可以试一下。

代码：

#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>
#define INF 0x3f3f3f3f
#define ll long long
const int N=10000005;	//18
const int MOD=1000;
using namespace std;
bool prime[N];
int p[N/10],pn;
void get_prime(){
	pn=0;
	memset(prime,false,sizeof(prime));
	prime[0]=prime[1]=true;
	for(long long i=2;i<N;i++){
		if(!prime[i]){
			p[pn++]=i;
			for(long long j=i*i;j<N;j+=i){
				prime[j]=true;
			}
		}
	}
}
int main(){
	get_prime();
	int T,n,num=1,ans;
	scanf("%d",&T);
	while(T--){
		scanf("%d",&n);
		ans=0;
		for(int i=0;p[i]<=n/2;i++){
			if(prime[n-p[i]]==false) ans++;
		}
		printf("Case %d: %d\n",num++,ans);
	}
	return 0;
}