hdu 1397 Goldbach's Conjecture

 题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1397

Goldbach's Conjecture

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6729    Accepted Submission(s): 2619

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

610120

 

Sample Output

121

 

Source

Asia 1998, Tokyo (Japan) 

题意：给一个合数n（不是素数的数），问能有多少种质数a，b组合的方式满足(a+b)=n(a<=b)

直接打素数表，然后模拟枚举即可，埃氏筛法和线性筛法均可

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<map>#include<set>#include<stack>#include<cmath>#include<queue>#include<deque>#include<utility>using namespace std;typedef long long ll;#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)#define mem(i,a) memset(i,a,sizeof i)#define sf scanf#define pf printfconst int Max=1e7+5;int p[Max/10],m,n,cou,T;//1e7范围内素数的个数只有1/10不到，如果题目卡内存的话，不能直接写成p[Max]bool vis[Max+1];//记录当前状态，false为素数，true不是素数void prime()//素数线性打表{    mem(vis,0);    cou=0;    for(int i=2;i<=Max;i++)    {        if(!vis[i])p[cou++]=i;        for(int j=0;j<cou&&i*p[j]<=Max;j++)        {            vis[i*p[j]]=1;            if(!(i%p[j]))break;        }    }}int main(){    prime();    sf("%d",&T);    for1(i,T)    {        sf("%d",&m);        ll ans=0;        for(int j=0;p[j]<=(m>>1);j++)//判断两个数是否同时为素数            if(!vis[m-p[j]])                ans++;        pf("Case %d: %lld\n",i,ans);    }    return 0;}