翻译哥德巴赫猜想

[zhanghuizi123]

请问 用C语言翻译哥德巴赫猜想

[zishare]

什么意思？
任何一个偶数，可以分解为两个质数之和

[尘缘udbwcso]

用JAVA做过

package prime;

public class Prime {
	//判断num是不是素数,是则返回true
	public boolean isPrime(int num){
		if(num == 2 || num == 3)
			return true;
		if(num%2 == 0)//num为偶数
			return false;
		
		//num为奇数
		int d = 3;
		while(d <= Math.sqrt(num) && num % d != 0){
			d += 2;
		}
		if(num % d != 0)
			return true;
		return false;		
	}

}
package prime;

/**
 * 任何大于２的偶数均为两个素数之和
 * @author Administrator
 *
 */

public class Goldbach {
	private int count;
	
	public void setCount(int count){
		this.count = count;
	}
	public int getCount(){
		return count;
	}
	public void inc(){
		count++;
	}
	//public void output(){
		//System.out.println(count);
	//}	
	
	
	//判断是不是两个素数之和，如果是则输出并返回true
	public boolean isSumOfPrime(int even){
		boolean sumOfPrime = false;
		Prime prime = new Prime();
		int factor1 = 0;
		int factor2 = 0;
		for(int i = 2; i <= even/2; i++){
			factor1 = i;
			factor2 = even - i;
			if(prime.isPrime(factor1) && prime.isPrime(factor2)){
				System.out.println(even + " = " + factor1 + " + " + factor2);
				sumOfPrime = true;
				break;
			}
		}
		return sumOfPrime;
	}
	public static void main(String[] args){
		Goldbach num = new Goldbach();
		num.setCount(0);
		for(int i = 4; i <= 10000; i += 2){
			if(num.isSumOfPrime(i))
				num.inc();
		}
		//num.output();
	}

}

[尘缘udbwcso]

因为刚学,写的比较差,
其实只能验证一定范围的数是否满足

[sailshiny]

有点深奥哦！

[qq149850508]

while(d <= sqrt((double)num) && num % d != 0)
{
d += 2;
}
5楼这个加上重载类型，不然（VS2010）编译不过

[尘缘udbwcso]

/**
 *任何大于２的偶数均为两个素数之和
 *验证4到10000
 */
#include <stdio.h>
#include <math.h>
//判断num是不是素数,是则返回1
int isPrime(int num)
{
	int d = 3;
	if(num == 2 || num == 3)
		return 1;
	if(num%2 == 0)//num为偶数
		return 0;
	//num为奇数
	while(d <= sqrt(num) && num % d != 0)
	{
		d += 2;
	}
	if(num % d != 0)
		return 1;
	return 0;
}

//判断是不是两个素数之和，如果是则输出并返回1
int isSumOfPrime(int even)
{
	int sumOfPrime = 0;
	int factor1 = 0;
	int factor2 = 0;
	for(int i = 2; i <= even/2; i++)
	{
		factor1 = i;
		factor2 = even - i;
		if(isPrime(factor1)==1 && isPrime(factor2)==1)
		{
			printf("%d = %d + %d\n", even, factor1, factor2);
			sumOfPrime = 1;
			break;
		}        
    }
	return sumOfPrime;
}

int main()
{
	int i;
	for(i = 4; i <= 10000; i += 2)
	{
            isSumOfPrime(i);
	}
	return 0;
}