lightoj 1259 Goldbach`s Conjecture

解题思路：按照常规暴力的话，结果肯定是Memory  Limit，所以这道题可以用素数筛先把素数找出来，记录在一个数组里面，等用的时候直接可以用。见代码吧！！！

这道题还有一个需要注意到，开数组的时候用bool型，因为bool型占一个字节，而int占四个字节

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, ...

#include<stdio.h>#include<string.h>#include<vector>using namespace std;bool vis[10000005];vector<int> v;void get_prime(){    memset(vis,false,sizeof(vis));    vis[1]=true;    for(int i = 2; i<=10000000; i++)    {        int t=10000000/i;        for(int j=2;j<=t;j++)            vis[i*j]=true;    }    for(int i=2;i<=10000000;i++)        if(!vis[i])        v.push_back(i);}int main(){    get_prime();    int T,t=1;    scanf("%d",&T);    while(T--)    {        int n,ans=0;        scanf("%d",&n);        for(int i=0;v[i]<=n/2;i++)        {            if(!vis[n-v[i]])                ans++;        }        printf("Case %d: %d\n",t++,ans);    }    return 0;}