Goldbach`s Conjecture

题目描述：

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 10^7.

输入：

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 ≤ 10^7, n is even).

输出：

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

样例输入：

2

6

4

样例输出：

Case 1: 1

Case 2: 1

题目大意：

给出几组测试数据，每组给出一个n，问n能被分成几对素数的和。

思路：先进行一个素数打表，把数据范围内所有素数存在一个数字内，当然此时已经是从小到大排好的了，然后从数组中的第一个1到最后一个遍历，如果n减去该元素的值还是一个素数的话num++，如果该元素大于等于n/2+1，结束遍历。输出num的值即可。

code：

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int maxn=1e7+5;
bool visit[maxn];
int prime[1000000];
int cnt=0;
int main()
{
	for(int i=2;i<=10000000; i++)
    {
        if(visit[i]==false)
        {
            prime[cnt++]=i;
            for(int j=i+i; j<=10000000; j=j+i)
                visit[j]=true;
        }
    }
    visit[0]=visit[1]=true;

	int tt;
	scanf("%d",&tt);
	for(int kk=1;kk<=tt;kk++)
	{
		int n;
		scanf("%d",&n);
		int sum=0;
		for(int i=0;i<cnt;i++)
		{
			if(prime[i]>=n/2+1)break;
			if(visit[n-prime[i]]==false)sum++;
		}//直接使用visit会超时
		printf("Case %d: %d\n",kk,sum);
	}
	return 0;
 }