Goldbach`s Conjecture LightOJ - 1259（素数打表）

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.
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 ≤ 10^7, 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：

给定一个n，判断符合条件a+b=n且a,b均为素数且a<=b的a，b方案有多少种。

思路：

先对1e7范围内素数打表，然后判断相加是否等于0 就可，可以看下面代码。

代码：

#include
bool prime[10000010];
void is_prime()
{
    int i,j;
    for(i=2; i*i<=10000000; i++)
    {
        if(prime[i]==0)
        {
            for(j=i*i; j<=10000000; j+=i)
                prime[j]=1;
        }
    }
}
int main()
{
    is_prime();
    int t;
    scanf("%d",&t);
    int n,i,j,k1=0;

    while(t--)
    {
        k1++;
        scanf("%d",&n);
        int k=0;
        for(i=2; i<=n/2; i++)
        {
            if(prime[i]+prime[n-i]==0)
            {
                k++;
            }
        }
        printf("Case %d: %d\n",k1,k);
    }
    return 0;
}