题目1440：Goldbach's Conjecture

题目1440：Goldbach's Conjecture

时间限制：1 秒

内存限制：128 兆

特殊判题：否

题目描述：

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.

输入：

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.

输出：

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

样例输入：

6
10
12
0

样例输出：

1
2
1
解题思路：先找出一定范围内的素数，存起来，然后遍历，找到符合的话，count++。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<math.h>
using namespace std;
# define len 200000
int prime[len];//用来存储素数
int primesize; //素数个数
bool mark[len];//标记一个数是不是素数
 
void init()
{
    primesize=0;//素数个数初始化为0
    long long int i,j;
    for( i=0; i<len; i++)
    {
        mark[i]=false;//为true时，说明其不是素数
    }
    for( i=2; i<len; i++)
    {
        if(mark[i]==true)  continue;//不是素数继续
        else
        {
            prime[primesize++]=i;
            for( j=i*i; j<len; j+=i)
            {
                mark[j]=true;
            }
        }
    }
}
 
 
int main()
{
    init();
    int n;
 
    while(scanf("%d",&n)!=EOF&&n)
    {
            int count=0;
        for(int i=0; i<primesize&&prime[i]<n; i++)
            for(int j=i; j<primesize&&prime[j]<n; j++)
            {
                if(prime[i]+prime[j]==n)
                    count++;
            }
        printf("%d\n",count);
    }
    return 0;
}
 
/**************************************************************
    Problem: 1440
    User: zhuoyuezai
    Language: C++
    Result: Accepted
    Time:70 ms
    Memory:2496 kb
****************************************************************/

 方法2，不是先生成一个素数表，而是暴利判断，貌似时间短了一些

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#include<math.h>
using namespace std;
 
bool judge(int n)
{
    if(n<=1)
        return false;
    int bound =(int)(sqrt(n)+1);
    for(int i=2; i<bound; i++)
    {
        if(n%i==0)
            return false;
    }
    return true;
 
}
 
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n)
    {
        int count=0;
        int m=n/2;
        for(int i=2; i<=m; i++)
        {
            if(judge(i)&&judge(n-i))
                count++;
        }
        printf("%d\n",count);
    }
    return 0;
}
 
/**************************************************************
    Problem: 1440
    User: zhuoyuezai
    Language: C++
    Result: Accepted
    Time:20 ms
    Memory:1532 kb
****************************************************************/