【哥德巴赫猜想】LightOJ Goldbach`s Conjecture 1259_lightoj

本文链接：https://blog.youkuaiyun.com/ydd97/article/details/48260901

本文探讨了哥德巴赫猜想，并提供了一种算法来验证该猜想对于小于等于10^7的所有偶数是否成立。通过预先计算素数并使用高效的数据结构，实现了快速检查任意给定整数是否为两个素数之和。

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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

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：

一个数可以有多少种可能是由两个素数相加得到的。

解题思路：

哥德巴赫猜想，此题需要素数打表，还要注意打表后的判断方法，不能再采用普通的枚举算法，会超时。数组一定要开成bool型的，否则超内存。

AC代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>

using namespace std;

int prime[700000];
bool is_prime[10000000];
int tot;

void solve()
{
    tot=0;
    is_prime[0]=is_prime[1]=1;
    for(int i=2;i<10000000;i++)
        if(!is_prime[i]){
            prime[tot++]=i;
            for(int j=i*2;j<10000000;j+=i){
                is_prime[j]=1;
            }
        }
}

int main()
{
    solve();
    int t;
    int xp=0;
    scanf("%d",&t);
    while(t--){
        int n;
        scanf("%d",&n);
        int res=0;
        for(int i=0;prime[i]<=n/2;i++){
            int j=n-prime[i];
            if(!is_prime[j]) res++;
        }
        printf("Case %d: %d\n",++xp,res);
    }
    return 0;
}