LightOJ-1259-Goldbach`s Conjecture-素数打表+判断素数对数

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：给出n个数，判断每个数由几对素数组成

 

• 注意：pri开数组的范围

　　　　　 打表标记用bool，不然一直RT

　　　　　 判断素数对数当走到n/2时后面肯定找不到一对了，这个时候可以break了

 

判断有几对素数：

for(int i=0; i<p; i++)
{
    if(pri[i]>n/2)
        break;
    if(book[n-pri[i]]==0)
        ans++;
}

 

 

#include<stdio.h>
#include<cmath>
#include<algorithm>
#include<string.h>
typedef long long ll;
using namespace std;

//const int N=1e7+20;
bool book[10000001];
int pri[5000000];
int p;

void prime()
{
    //非素数标记为1
    //memset(book,0,sizeof(book));
    // memset(pri,0,sizeof(pri));这里自己会清空，不要随便去清空，耗时
    book[0]=1;
    book[1]=1;
    p=0;
    for(int i=2; i<=10000000; i++)
    {
        if(book[i]==0)
        {
            pri[p++]=i;//记录素数元素
            for(int j=i*2; j<=10000000; j+=i)
                book[j]=1;
        }
    }
}

int main()
{
    prime();
    // sort(pri,pri+p);
    int t,tt=1;
    scanf("%d",&t);
    int n;
    while(t--)
    {
        scanf("%d",&n);
        int ans=0;
43         for(int i=0; i<p; i++)
44         {
45             if(pri[i]>n/2)
46                 break;
47             if(book[n-pri[i]]==0)
48                 ans++;
49         }
        printf("Case %d: %d\n",tt++,ans);
    }
    return 0;
}