Goldbach`s Conjecture LightOJ 1259

题目链接

Goldbach`s Conjecture

Description

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

解题思路：这个题非常坑，题上显示数据范围(10^7),使用int数组标记筛选,内存超限，需要使用bool型数组，因为int型每个数字占四个字节，而bool型占一个字节。还有一点，需要把素数存入一个向量里面（用int 数组存，编译器会出现错误，不知道为什么？）。

#include <stdio.h>#include <string.h>#define maxn 10000000+1#include <vector>using namespace std;bool num[maxn];vector<int>Q;int n;void prime(){    memset(num,false,sizeof(num));    for(int i=2; i<=maxn; i++)    {        if(num[i]==false)        {            for(int j=i+i; j<=maxn; j+=i)                num[j]=true;        }    }    Q.push_back(2);    for(int i=3; i<=maxn; i+=2)        if(num[i]==false)            Q.push_back(i);}int main(){    int t,Case=1;    prime();    scanf("%d",&t);    while(t--)    {        int n,cont=0;        scanf("%d",&n);        for(int i=0; Q[i]<=n/2; i++)        {            if(num[n-Q[i]]==false)                cont++;        }        printf("Case %d: %d\n",Case++,cont);    }    return 0;}