poj2262 筛选法求素数

最新推荐文章于 2021-02-02 10:09:10 发布

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本文介绍了一种使用筛选法求解素数的方法，并通过一个具体示例验证了哥德巴赫猜想对于小于一百万的所有偶数的有效性。程序通过高效筛选素数并找出合适的素数对来完成任务。

Goldbach's Conjecture

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28727		Accepted: 11007

Description

In 1742, Christian Goldbach, a German amateur mathematician, sent a letter to Leonhard Euler in which he made the following conjecture:

Every even number greater than 4 can be
written as the sum of two odd prime numbers.

For example:

8 = 3 + 5. Both 3 and 5 are odd prime numbers.
20 = 3 + 17 = 7 + 13.
42 = 5 + 37 = 11 + 31 = 13 + 29 = 19 + 23.

Today it is still unproven whether the conjecture is right. (Oh wait, I have the proof of course, but it is too long to write it on the margin of this page.)
Anyway, your task is now to verify Goldbach's conjecture for all even numbers less than a million.

Input

The input will contain one or more test cases.
Each test case consists of one even integer n with 6 <= n < 1000000.
Input will be terminated by a value of 0 for n.

Output

For each test case, print one line of the form n = a + b, where a and b are odd primes. Numbers and operators should be separated by exactly one blank like in the sample output below. If there is more than one pair of odd primes adding up to n, choose the pair where the difference b - a is maximized. If there is no such pair, print a line saying "Goldbach's conjecture is wrong."

Sample Input

Sample Output

8 = 3 + 5
20 = 3 + 17
42 = 5 + 37

#include<iostream>
#include<string.h>
#include<stdio.h>
#define MAX 1000000
using namespace std;
bool prime1[MAX+1];
int prime2[MAX/2];
int  cal()
{
    prime1[1]=0;prime1[2]=1;
    for(int i=3;i<=MAX;i++)
    {
        if(i%2)prime1[i]=1;
        else prime1[i]=0;
    }
    int count=0;
    for(int i=3;i<=MAX;i++)
    {
        if(prime1[i])
        {
            if(i*i>MAX)break;
            for(int j=i*i;j<=MAX;j+=i)
            {
                prime1[j]=0;
            }
        }
    }
    for(int i=1;i<=MAX;i++)
    {
        if(prime1[i])
        {
            prime2[++count]=i;
        }
    }
    return count;
}
int main()
{
    int count=cal();
    int n;
    while(cin>>n,n)
    {
        bool temp=true;
        for(int i=1;i<=count;i++)
        {
            if(prime2[i]>n)break;
            if(prime1[n-prime2[i]])
            {
                temp=false;
                printf("%d = %d + %d\n",n,prime2[i],n-prime2[i]);
                break;
            }
        }
        if(temp)
        {
            cout<<"Goldbach's conjecture is wrong."<<endl;
        }
    }
    return 0;
}

这个问题是一个筛选法求素数的问题。。第一次做我超时了，主要是在找利用两个素数因子的时候用了一个两层循环没有好好利用prime1.。。。