LightOJ - 1259

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

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:给出一个数，求两素素相加等于它的情况总和。

思路:素数打个表，直接暴力就能过

               

#include<stdio.h>
#include<algorithm>
#include<string.h>
#define M 10000005
using namespace std;
bool b[M+5];
int a[664579+10];
int k=0;
void init()
{
    for(int i=2;i<M;i++)
    {
        if(!b[i])
        {
            a[k++]=i;
            for(int j=i+i;j<M;j+=i)
            {
                b[j]=1;
            }
        }
    }
}
int erfen(int n)
{
    int l=0,r=k,pan;
    while(l<=r)
    {
        pan=(l+r)/2;
        if(a[pan]>=n)
            r=pan-1;
        else
            l=pan+1;
    }
    return a[l]==n;
}
int main()
{
    init();
    int t,n,ph=1,ans;
    scanf("%d",&t);
    while(t--)
    {
        ans=0;
        scanf("%d",&n);
        for(int i=0;i<k&&a[i]<=n/2;i++)
        {
//            if(erfen(n-a[i]))//事实证明，用数组比用二分快的多
            if(!b[n-a[i]])
            {
                ans++;
            }
        }
        printf("Case %d: %d\n",ph++,ans);
    }
}