hdoj 1397 Goldbach's Conjecture （大数素数打表）

最新推荐文章于 2018-01-07 12:41:17 发布

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最新推荐文章于 2018-01-07 12:41:17 发布

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分类专栏：大数文章标签： acm

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博客讲述了在解决hdoj 1397问题时，由于合数分解导致的TLE错误，从而采用素数打表的方法应对大数范围（2^15）内的素数判断。

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这道题本以为就是简单的合数分解，求素数组合数，没想到总是TLE，后来才发现里面的数范围特别大 2^15，所以只能打表了（把所有2^15以内素数标记为1，不是的标记为0）

Goldbach's Conjecture

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5320 Accepted Submission(s): 2043

Problem Description

Goldbach's Conjecture: For any even number n greater than or equal to 4, there exists at least one pair of prime numbers p1 and p2 such that n = p1 + p2.
This conjecture has not been proved nor refused yet. No one is sure whether this conjecture actually holds. However, one can find such a pair of prime numbers, if any, for a given even number. The problem here is to write a program that reports the number of all the pairs of prime numbers satisfying the condition in the conjecture for a given even number.

A sequence of even numbers is given as input. Corresponding to each number, the program should output the number of pairs mentioned above. Notice that we are interested in the number of essentially different pairs and therefore you should not count (p1, p2) and (p2, p1) separately as two different pairs.

Input

An integer is given in each input line. You may assume that each integer is even, and is greater than or equal to 4 and less than 2^15. The end of the input is indicated by a number 0.

Output

Each output line should contain an integer number. No other characters should appear in the output.

Sample Input

Sample Output

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<iostream>
using namespace std;
#define max (1<<15)+10
bool prime[max];
void fun()
{
	memset(prime,1,sizeof(prime));
	for(int i=2;i<max;i++)
		for(int j=2;j<=sqrt(i);j++)
		{
			if(i%j==0)
			{
				prime[i]=0;
				break;	
			}
		}
		prime[0]=prime[1]=0;
}
/*void calcPrime()

{
    memset(prime, true, sizeof(prime));   //另一种达标方法，我试了下，比自己的还慢 
    for(int i=2; i<max; i++)
    {
       for(int j=i*2; j<max; j+=i) 
	   prime[j] = false;
    }
    prime[0] = prime[1] = false;
}*/

int main()
{	
	fun(); //必须调用先把素数标记出来，后边才能判断 
	int m,n;
	while(scanf("%d",&n)&&n)
	{    m=0;
		/*for(int i=2;i<=n/2 ;i++)
		{
			if(prime[i] && prime[n-i])
			{
			  m++;	
			}  
		}*/
		 
		 for(int i=2; i<=n/2; i++)

            if(prime[i] && prime[n-i]) m++;
            
		printf("%d\n",m);
	}
	return 0;
}