Goldbach`s Conjecture LightOJ

Goldbach`s Conjecture   LightOJ - 1259

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

Hint

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, ...

思路：用筛选法打表，打表的为不是素数的。注意判断不能有重复的。

#include<stdio.h>#include<string.h>#define MAXN 10000000bool book[MAXN];int a[666666];int p=0;int pri(){    memset(book,false ,sizeof(book));

    book[0]=true;    book[1]=true;    for(int i=2;i<=MAXN;i++)    {        if(!book[i])        {            a[p++]=i;            for(int j=2*i;j<=MAXN;j+=i)                book[j]=true;        }    }}

int main(){    pri();    int t,n,g=0;

    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        int sum=0;        for(int i=0;i<p;i++)        {            if(a[i]>=n/2+1)                break;            if(!book[n-a[i]]&&n>=2*a[i])               sum++;        }        printf("Case %d: ",++g);        printf("%d\n",sum);    }}