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

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Sample Input

Sample Output

Case 1: 1

Case 2: 1

Note

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++;
}

 1 #include<stdio.h>
 2 #include<cmath>
 3 #include<algorithm>
 4 #include<string.h>
 5 typedef long long ll;
 6 using namespace std;
 7 
 8 //const int N=1e7+20;
 9 bool book[10000001];
10 int pri[5000000];
11 int p;
12 
13 void prime()
14 {
15     //非素数标记为1
16     //memset(book,0,sizeof(book));
17     // memset(pri,0,sizeof(pri));这里自己会清空，不要随便去清空，耗时
18     book[0]=1;
19     book[1]=1;
20     p=0;
21     for(int i=2; i<=10000000; i++)
22     {
23         if(book[i]==0)
24         {
25             pri[p++]=i;//记录素数元素
26             for(int j=i*2; j<=10000000; j+=i)
27                 book[j]=1;
28         }
29     }
30 }
31 
32 int main()
33 {
34     prime();
35     // sort(pri,pri+p);
36     int t,tt=1;
37     scanf("%d",&t);
38     int n;
39     while(t--)
40     {
41         scanf("%d",&n);
42         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         }
50         printf("Case %d: %d\n",tt++,ans);
51     }
52     return 0;
53 }

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

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