POJ 2262-Goldbach's Conjecture(素数筛)

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Description

Input

Output

Sample Input

8
20
42
0

Sample Output

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

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long LL;
int prime[1000010]={2,3,5};
int k=3;
int ip[1000010]={0};
void is_prime()
{
    int i,j;
    int flag=0;
    int gad=2;
    ip[2]=1;ip[3]=1;ip[5]=1;
    for(i=7;i<=1000010;i+=gad){
        flag=0;
        gad=6-gad;
        for(j=0;prime[j]*prime[j]<=i;j++){
            if(i%prime[j]==0){
                flag=1;
                break;
            }
        }
        if(!flag){
            prime[k++]=i;
            ip[i]=1;
        }
    }
}
int main()
{
    int n,i,j;
    is_prime();
    while(~scanf("%d",&n)){
        if(!n) break;
        int f=0;
        for(i=0;i<=k;i++){
            j=n-prime[i];
            if(ip[j]){
                f=1;
                break;
            }
        }
        if(!f)
            printf("Goldbach's conjecture is wrong.\n");
        else
            printf("%d = %d + %d\n",n,prime[i],j);

    }
    return 0;
}

POJ 2262-Goldbach's Conjecture(素数筛)

标签：

原文地址：http://blog.csdn.net/u013486414/article/details/44260305