LightOJ1259 Goldbach`s Conjecture —— 素数表

题目链接：https://vjudge.net/problem/LightOJ-1259

1259 - Goldbach`s Conjecture

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
2 6 4	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：

哥德巴赫猜想：任何一个大于2的偶数，都可以是两个素数的和。给出一个偶数，判断有多少对素数的和是这个数。

题解：

由于n<=1e7，所以我们可以先筛选出1e7范围内的素数，然后再枚举每一个素数进行判断。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 1e7+10;

bool notprime[MAXN];
int prime[1000010];
void getPrime()
{
    memset(notprime, false, sizeof(notprime));
    notprime[0] = notprime[1] = true;
    prime[0] = 0;
    for (int i = 2; i<=MAXN; i++)
    {
        if (!notprime[i])prime[++prime[0]] = i;
        for (int j = 1; j<=prime[0 ]&& prime[j]<=MAXN/i; j++)
        {
            notprime[prime[j]*i] = true;
            if (i%prime[j] == 0) break;
        }
    }
}

int main()
{
    getPrime();
    int T, n, kase = 0;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int ans = 0;
        for(int i = 1; prime[i]<=n/2; i++)
            if(!notprime[n-prime[i]])
                ans++;
        printf("Case %d: %d
", ++kase,ans);
    }
    return 0;
}