| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10364 | Accepted: 6143 |
Description
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. There can be many such numbers. 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 215. 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
6 10 12 0
Sample Output
1 2 1
Source
水题
写一遍的目的是。。。复习一下快速筛的写法 喵呜
1 /************************************************************************* 2 > File Name: code/poj/2909.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年08月22日 星期六 14时25分34秒 6 ************************************************************************/ 7 8 #include<iostream> 9 #include<iomanip> 10 #include<cstdio> 11 #include<algorithm> 12 #include<cmath> 13 #include<cstring> 14 #include<string> 15 #include<map> 16 #include<set> 17 #include<queue> 18 #include<vector> 19 #include<stack> 20 #define y0 abc111qqz 21 #define y1 hust111qqz 22 #define yn hez111qqz 23 #define j1 cute111qqz 24 #define tm crazy111qqz 25 #define lr dying111qqz 26 using namespace std; 27 #define REP(i, n) for (int i=0;i<int(n);++i) 28 typedef long long LL; 29 typedef unsigned long long ULL; 30 const int inf = 0x3f3f3f3f; 31 const int N=1<<16; 32 bool not_prime[N]; 33 int prime[N]; 34 int prime_num; 35 int n; 36 37 void init(){ 38 not_prime[0] = true; 39 not_prime[1] = true; 40 for ( int i =2 ; i < N ; i++){ 41 if (!not_prime[i]){ 42 prime[++prime_num] = i; 43 } 44 for ( int j = 1 ; j <= prime_num&&i*prime[j]<N ; j++){ 45 not_prime[i*prime[j]] = true; 46 if (i%prime[j]==0) break; 47 } 48 } 49 } 50 int main() 51 { 52 init(); 53 int n ; 54 while (scanf("%d",&n)&&n){ 55 int ans = 0 ; 56 for ( int i = 2 ; i <= n /2 ; i++){ 57 if (!not_prime[i]&&!not_prime[n-i]){ 58 ans++; 59 } 60 } 61 printf("%d ",ans); 62 } 63 64 return 0; 65 }